Mix Examples-Circle and System of Circles Questions in English

(A) Let the center of the circle be $(h, k)$.
The chord is given by the equation $5x + 2y = 16$,which can be rewritten as $5x + 2y - 16 = 0$.
The slope of this chord is $m_1 = -\frac{5}{2}$.
The line segment joining the center $(h, k)$ and the midpoint $(2, 3)$ of the chord is perpendicular to the chord.
The slope of this line segment is $m_2 = \frac{k - 3}{h - 2}$.
Since the line segment is perpendicular to the chord,$m_1 \times m_2 = -1$.
$-\frac{5}{2} \times \frac{k - 3}{h - 2} = -1$
$5(k - 3) = 2(h - 2)$
$5k - 15 = 2h - 4$
$2h - 5k + 11 = 0$
Replacing $(h, k)$ with $(x, y)$,the locus is $2x - 5y + 11 = 0$.

(B) The given circle is $x^2 + y^2 + 4x - 6y - 12 = 0$.
The center of the circle is $C(-2, 3)$ and the radius $R = \sqrt{(-2)^2 + 3^2 - (-12)} = \sqrt{4 + 9 + 12} = \sqrt{25} = 5$.
Let the chord subtend an angle of $\theta = \frac{\pi}{3} = 60^\circ$ at the circumference.
By the property of circles,the angle subtended by the chord at the center is $2\theta = 2 \times 60^\circ = 120^\circ$.
Let $(h, k)$ be the midpoint of the chord. The distance $p$ from the center $C(-2, 3)$ to the chord is given by $p = R \cos(\frac{2\theta}{2}) = R \cos(60^\circ)$.
$p = 5 \times \frac{1}{2} = 2.5$.
The distance from the center $(-2, 3)$ to the point $(h, k)$ is $p$,so the equation is $(h - (-2))^2 + (k - 3)^2 = p^2$.
$(h + 2)^2 + (k - 3)^2 = (2.5)^2 = 6.25$.
Replacing $(h, k)$ with $(x, y)$,the locus is $(x + 2)^2 + (y - 3)^2 = 6.25$.

(B) The centers of the circles are $C_1 = (1, 0)$ and $C_2 = (0, 2)$,and their radii are $r_1 = \sqrt{10}$ and $r_2 = \sqrt{5}$.
The distance between the centers $d$ is given by $d = \sqrt{(1 - 0)^2 + (0 - 2)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}$.
The angle of intersection $\theta$ between two circles is given by $\cos \theta = \frac{r_1^2 + r_2^2 - d^2}{2 r_1 r_2}$.
Substituting the values,we get $\cos \theta = \frac{10 + 5 - 5}{2 \cdot \sqrt{10} \cdot \sqrt{5}} = \frac{10}{2 \cdot \sqrt{50}} = \frac{10}{2 \cdot 5\sqrt{2}} = \frac{10}{10\sqrt{2}} = \frac{1}{\sqrt{2}}$.
Therefore,$\theta = \frac{\pi}{4}$.

(B) The given circle is $x^2 + y^2 + 2x - 1 = 0$.
Completing the square,we get $(x + 1)^2 + y^2 = 2$.
This represents a circle with center $(-1, 0)$ and radius $r = \sqrt{2}$.
The line is $x - y + c = 0$.
For the line to intersect the circle at exactly one point,the perpendicular distance from the center $(-1, 0)$ to the line must be equal to the radius.
Distance $d = \frac{|(-1) - (0) + c|}{\sqrt{1^2 + (-1)^2}} = \frac{|c - 1|}{\sqrt{2}}$.
Setting $d = r$,we have $\frac{|c - 1|}{\sqrt{2}} = \sqrt{2}$.
$|c - 1| = 2$.
This gives $c - 1 = 2$ or $c - 1 = -2$.
Therefore,$c = 3$ or $c = -1$.

(C) Let the equation of the two circles be $(x-r)^2 + (y-r)^2 = r^2$ (since they touch both axes,the center is $(r, r)$ and the radius is $r$).
This simplifies to $x^2 + y^2 - 2rx - 2ry + r^2 = 0$.
Let the radii of the two circles be $r_1$ and $r_2$. The condition for two circles $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$ to cut orthogonally is $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.
Here,$g_1 = g_2 = -r_1, -r_2$ and $f_1 = f_2 = -r_1, -r_2$ and $c_1 = r_1^2, c_2 = r_2^2$.
So,$2(-r_1)(-r_2) + 2(-r_1)(-r_2) = r_1^2 + r_2^2$,which gives $4r_1r_2 = r_1^2 + r_2^2$.
Since the circles pass through $(a, b)$,we have $a^2 + b^2 - 2r(a+b) + r^2 = 0$,or $r^2 - 2r(a+b) + (a^2+b^2) = 0$.
For this quadratic in $r$,the sum of roots $r_1 + r_2 = 2(a+b)$ and the product of roots $r_1r_2 = a^2 + b^2$.
We know $r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2r_1r_2$.
Substituting into the orthogonality condition: $4r_1r_2 = (r_1 + r_2)^2 - 2r_1r_2$,so $6r_1r_2 = (r_1 + r_2)^2$.
Substituting the values: $6(a^2 + b^2) = [2(a+b)]^2 = 4(a^2 + b^2 + 2ab)$.
$6a^2 + 6b^2 = 4a^2 + 4b^2 + 8ab$.
$2a^2 - 8ab + 2b^2 = 0$,which simplifies to $a^2 - 4ab + b^2 = 0$.

(D) Let the circle be $x^2 + y^2 = 1$. The radius $r = 1$ and the center is $(0, 0)$.
Let the point be $P(a, 0)$. The distance from the center to $P$ is $d = |a|$.
Let $\theta$ be the angle between the tangents. Then $\sin(\theta/2) = \frac{r}{d} = \frac{1}{|a|}$.
Given $\frac{\pi}{2} < \theta < \pi$,we have $\frac{\pi}{4} < \frac{\theta}{2} < \frac{\pi}{2}$.
Taking the sine function,since $\sin(x)$ is increasing in this interval,we get $\sin(\frac{\pi}{4}) < \sin(\frac{\theta}{2}) < \sin(\frac{\pi}{2})$.
This implies $\frac{1}{\sqrt{2}} < \frac{1}{|a|} < 1$.
Taking the reciprocal,we get $1 < |a| < \sqrt{2}$.
This means $|a| > 1$ and $|a| < \sqrt{2}$.
Thus,$a \in (-\sqrt{2}, -1) \cup (1, \sqrt{2})$.

(C) Let the centre of the circle be $C(h, 0)$ since it lies on the $x$-axis.
Since the circle touches the line $x + y = 0$ at $P(2, -2)$,the radius $r$ is the distance $CP$.
$r^2 = (h - 2)^2 + (0 - (-2))^2 = (h - 2)^2 + 4$.
The line $CP$ is perpendicular to the tangent $x + y = 0$ (slope $-1$).
Thus,the slope of $CP = \frac{0 - (-2)}{h - 2} = \frac{2}{h - 2}$.
Since $CP \perp$ tangent,$(\frac{2}{h - 2}) \times (-1) = -1$ $\Rightarrow \frac{2}{h - 2} = 1$ $\Rightarrow h - 2 = 2$ $\Rightarrow h = 4$.
The centre is $C(4, 0)$.
$r^2 = (4 - 2)^2 + 4 = 4 + 4 = 8$,so $r = \sqrt{8} = 2\sqrt{2}$.
The equation of the circle is $(x - 4)^2 + y^2 = 8$.
The maximum value of $\alpha$ (the $x$-coordinate) is $h + r = 4 + 2\sqrt{2}$.

(C) Let the radii of the three concentric circles be $r_1, r_2, r_3$ in $A.P.$ such that $r_3 = 1$ (the largest). Let the common difference be $d > 0$. Then $r_3 = 1$,$r_2 = 1 - d$,and $r_1 = 1 - 2d$.
For the line $x - y + 1 = 0$ to cut all circles at two real and distinct points,the perpendicular distance from the center $(0, 0)$ to the line must be less than the radius of the smallest circle $r_1$.
The perpendicular distance $p = \frac{|0 - 0 + 1|}{\sqrt{1^2 + (-1)^2}} = \frac{1}{\sqrt{2}}$.
Thus,we require $r_1 > p$,which means $1 - 2d > \frac{1}{\sqrt{2}}$.
Solving for $d$: $2d < 1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2}}$.
$d < \frac{\sqrt{2} - 1}{2\sqrt{2}} = \frac{2 - \sqrt{2}}{4}$.
Since $d > 0$,the interval for $d$ is $\left( 0, \frac{2 - \sqrt{2}}{4} \right)$.

(B) Let the point on the circle $x^2 + y^2 = a^2$ be $(a \cos \theta, a \sin \theta)$.
The equation of the chord of contact of tangents drawn from $(a \cos \theta, a \sin \theta)$ to the circle $x^2 + y^2 = b^2$ is given by $x(a \cos \theta) + y(a \sin \theta) = b^2$.
This line touches the circle $x^2 + y^2 = c^2$.
The perpendicular distance from the center $(0, 0)$ to the line $x(a \cos \theta) + y(a \sin \theta) - b^2 = 0$ must be equal to the radius $c$.
So,$\frac{|-b^2|}{\sqrt{(a \cos \theta)^2 + (a \sin \theta)^2}} = c$.
$\frac{b^2}{\sqrt{a^2(\cos^2 \theta + \sin^2 \theta)}} = c$.
$\frac{b^2}{a} = c$,which implies $b^2 = ac$.
Since $b^2 = ac$,the terms $a, b, c$ are in $G.P.$

(B) Both circles pass through the origin $(0, 0)$.
Let the centers be $C_1 = (-g_1, -f_1)$ and $C_2 = (-g_2, -f_2)$.
Since the circles touch each other at the origin $(0, 0)$,the centers $C_1$,$C_2$,and the point of contact $(0, 0)$ must be collinear.
The slope of the line joining $C_1$ and $(0, 0)$ is $\frac{-f_1 - 0}{-g_1 - 0} = \frac{f_1}{g_1}$.
The slope of the line joining $C_2$ and $(0, 0)$ is $\frac{-f_2 - 0}{-g_2 - 0} = \frac{f_2}{g_2}$.
Since the points are collinear,the slopes must be equal:
$\frac{f_1}{g_1} = \frac{f_2}{g_2}$.

(A) Let the point of intersection of the tangents be $(x_1, y_1)$. The chord of contact of the tangents from $(x_1, y_1)$ to the circle $x^2 + y^2 = 1$ is given by $xx_1 + yy_1 = 1$,or $xx_1 + yy_1 - 1 = 0$ $... (1)$.
The common chord of the two circles $S_1: x^2 + y^2 - 1 = 0$ and $S_2: x^2 + y^2 - (\lambda + 6)x + (8 - 2\lambda)y - 3 = 0$ is given by $S_1 - S_2 = 0$.
$(x^2 + y^2 - 1) - (x^2 + y^2 - (\lambda + 6)x + (8 - 2\lambda)y - 3) = 0$
$(\lambda + 6)x - (8 - 2\lambda)y + 2 = 0$ $... (2)$.
Since the chord of contact is the same as the common chord,we compare equations $(1)$ and $(2)$:
$\frac{x_1}{\lambda + 6} = \frac{y_1}{-(8 - 2\lambda)} = \frac{-1}{2}$
From $\frac{x_1}{\lambda + 6} = -\frac{1}{2}$,we get $\lambda + 6 = -2x_1 \Rightarrow \lambda = -2x_1 - 6$.
From $\frac{y_1}{2\lambda - 8} = -\frac{1}{2}$,we get $2\lambda - 8 = -2y_1 \Rightarrow \lambda = -y_1 + 4$.
Equating the two expressions for $\lambda$:
$-2x_1 - 6 = -y_1 + 4$
$y_1 - 2x_1 = 10$
$2x_1 - y_1 + 10 = 0$.
Replacing $(x_1, y_1)$ with $(x, y)$,the locus is $2x - y + 10 = 0$.

(A) Let the coordinates of vertex $A$ be $(a, b)$ and the centroid $G$ be $(h, k)$.
Since $\angle BAC = 90^o$ and $B(3, 0), C(-3, 0)$,the point $A$ lies on a circle with diameter $BC$. The center of this circle is the origin $(0, 0)$ and the radius is $3$. Thus,the equation of the circle is $x^2 + y^2 = 3^2 = 9$. Therefore,$a^2 + b^2 = 9$.
The centroid $G(h, k)$ of $\Delta ABC$ is given by $h = \frac{a + 3 - 3}{3} = \frac{a}{3}$ and $k = \frac{b + 0 + 0}{3} = \frac{b}{3}$.
This gives $a = 3h$ and $b = 3k$.
Substituting these into the circle equation $a^2 + b^2 = 9$,we get $(3h)^2 + (3k)^2 = 9$.
$9h^2 + 9k^2 = 9$,which simplifies to $h^2 + k^2 = 1$.
Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 = 1$.

(C) The equation of the circle is $x^2 + y^2 - ax - \frac{b}{2}y = 0$.
Let the chord pass through $P(a, b/2)$ and be bisected by the $x$-axis at some point $Q(h, 0)$.
The midpoint of the chord is $Q(h, 0)$. The line segment $PQ$ is a chord of the circle.
The center of the circle is $C(a/2, b/4)$.
The line $CQ$ must be perpendicular to the chord $PQ$.
The slope of $PQ$ is $m_1 = \frac{b/2 - 0}{a - h} = \frac{b}{2(a - h)}$.
The slope of $CQ$ is $m_2 = \frac{0 - b/4}{h - a/2} = \frac{-b/4}{h - a/2} = \frac{-b}{4h - 2a}$.
Since $PQ \perp CQ$,$m_1 \times m_2 = -1$.
$\frac{b}{2(a - h)} \times \frac{-b}{4h - 2a} = -1$
$\frac{-b^2}{2(a - h) \times 2(2h - a)} = -1$
$b^2 = 4(a - h)(2h - a) = 4(2ah - a^2 - 2h^2 + ah) = 4(-2h^2 + 3ah - a^2) = -8h^2 + 12ah - 4a^2$.
$8h^2 - 12ah + 4a^2 + b^2 = 0$.
For two distinct chords to exist,there must be two distinct values of $h$.
Thus,the discriminant $D > 0$.
$D = (-12a)^2 - 4(8)(4a^2 + b^2) > 0$
$144a^2 - 32(4a^2 + b^2) > 0$
$144a^2 - 128a^2 - 32b^2 > 0$
$16a^2 > 32b^2$
$a^2 > 2b^2$.

(C) Let the point on the line be $P(x_1, y_1)$. Since $P$ lies on the line $2x + y = 4$,we have $2x_1 + y_1 = 4$ ... $(1)$.
The equation of the chord of contact of tangents drawn from $P(x_1, y_1)$ to the unit circle $x^2 + y^2 = 1$ is $xx_1 + yy_1 = 1$.
Let $(h, k)$ be the midpoint of this chord. The equation of a chord with midpoint $(h, k)$ for the circle $x^2 + y^2 = 1$ is $xh + yk = h^2 + k^2$.
Comparing the two equations of the same chord:
$\frac{x_1}{h} = \frac{y_1}{k} = \frac{1}{h^2 + k^2}$
Thus,$x_1 = \frac{h}{h^2 + k^2}$ and $y_1 = \frac{k}{h^2 + k^2}$.
Substituting these into equation $(1)$:
$2 \left( \frac{h}{h^2 + k^2} \right) + \frac{k}{h^2 + k^2} = 4$
$2h + k = 4(h^2 + k^2)$
Replacing $(h, k)$ with $(x, y)$,the locus is $4(x^2 + y^2) = 2x + y$.

(C) Let the common chord be $AB$ and let it intersect the line joining the centres $O_1$ and $O_2$ at $M$. Let $PM = h$ be half the length of the common chord.
In $\triangle O_1PM$,$\angle PO_1M = 90^\circ / 2 = 45^\circ$. Thus,$h = O_1M \tan 45^\circ = O_1M$.
In $\triangle O_2PM$,$\angle PO_2M = 60^\circ / 2 = 30^\circ$. Thus,$h = O_2M \tan 30^\circ = O_2M / \sqrt{3}$,which implies $O_2M = h\sqrt{3}$.
The distance between the centres is $O_1M + O_2M = h + h\sqrt{3} = h(1 + \sqrt{3})$.
Given $O_1O_2 = \sqrt{3} + 1$,we have $h(1 + \sqrt{3}) = \sqrt{3} + 1$,so $h = 1$.
The radius $r_1$ of $c_1$ is $O_1P = h / \sin 45^\circ = 1 / (1/\sqrt{2}) = \sqrt{2}$.
The radius $r_2$ of $c_2$ is $O_2P = h / \sin 30^\circ = 1 / (1/2) = 2$.
Therefore,the radii are $\sqrt{2}$ and $2$.

(D) Let the centers of the circles be $C_1, C_2, C_3$ with radii $r_1=3, r_2=3, r_3=1$. The sides of the triangle $C_1C_2C_3$ are $C_1C_2 = 3+3=6$,$C_1C_3 = 3+1=4$,and $C_2C_3 = 3+1=4$.
This is an isosceles triangle. Let $M$ be the midpoint of $C_1C_2$. The height $h = C_3M = \sqrt{4^2 - 3^2} = \sqrt{7}$.
Points $A$ and $B$ lie on $C_1C_3$ and $C_2C_3$ respectively. By similar triangles,the side length $AB$ of triangle $ABC$ is $\frac{AB}{C_1C_2} = \frac{C_3A}{C_3C_1} = \frac{4-3}{4} = \frac{1}{4}$. Thus,$AB = \frac{1}{4} \times 6 = \frac{3}{2}$.
The height of triangle $ABC$ from $C$ to $AB$ is $h' = h - (\text{distance of } C \text{ from } AB) = \sqrt{7} - \frac{3}{4}\sqrt{7} = \frac{\sqrt{7}}{4}$.
Area of $\triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{3}{2} \times \frac{\sqrt{7}}{4} = \frac{3\sqrt{7}}{16}$.

(B) Let the circle be $x^2 + y^2 = 1$,so the radius $r = 1$.
Let the tangents intersect at point $A$. The angle between the tangents is $60^o$,so the angle between each tangent and the line joining the origin to $A$ is $30^o$.
Let $O$ be the origin and $P$ be the point of tangency. In $\triangle OPA$,$\angle OPA = 90^o$ and $\angle OAP = 30^o$.
Thus,$PA = r \cot(30^o) = 1 \cdot \sqrt{3} = \sqrt{3}$.
The area of the quadrilateral formed by the origin,the two points of tangency,and $A$ is $2 \times (\text{Area of } \triangle OPA) = 2 \times (\frac{1}{2} \cdot PA \cdot r) = 2 \times (\frac{1}{2} \cdot \sqrt{3} \cdot 1) = \sqrt{3}$.
The angle subtended by the two points of tangency at the center is $180^o - 60^o = 120^o = \frac{2\pi}{3}$ radians.
The area of the circular sector is $\frac{1}{2} r^2 \theta = \frac{1}{2} \cdot 1^2 \cdot \frac{2\pi}{3} = \frac{\pi}{3}$.
The area enclosed by the tangents and the arc is the area of the quadrilateral minus the area of the sector: $\sqrt{3} - \frac{\pi}{3}$.

(D) For the circle $x^2 + y^2 = 4$,center $C_1 = (0, 0)$ and radius $r_1 = 2$.
For the circle $x^2 + y^2 - 6 \sqrt{3} x - 6y + 20 = 0$,center $C_2 = (3 \sqrt{3}, 3)$ and radius $r_2 = \sqrt{(3 \sqrt{3})^2 + 3^2 - 20} = \sqrt{27 + 9 - 20} = \sqrt{16} = 4$.
The distance between centers $C_1C_2 = \sqrt{(3 \sqrt{3} - 0)^2 + (3 - 0)^2} = \sqrt{27 + 9} = 6$.
Since $C_1C_2 = r_1 + r_2 = 2 + 4 = 6$,the circles touch each other externally.
The common tangent at the point of contact is the radical axis,given by $S_1 - S_2 = 0$.
$(x^2 + y^2 - 4) - (x^2 + y^2 - 6 \sqrt{3} x - 6y + 20) = 0$ $\Rightarrow 6 \sqrt{3} x + 6y - 24 = 0$ $\Rightarrow \sqrt{3} x + y = 4$.
Dividing by $2$,we get $\frac{\sqrt{3}}{2} x + \frac{1}{2} y = 2$.
Comparing this with $x \cos \theta + y \sin \theta = 2$,we have $\cos \theta = \frac{\sqrt{3}}{2}$ and $\sin \theta = \frac{1}{2}$.
Thus,$\theta = \frac{\pi}{6}$.

(C) Let the radius of the smaller circle be $r_1 = 2$ and the radius of the larger circle be $r_2$.
According to the problem,the area of the smaller circle is equal to the area of the region between the two circles.
$\pi r_1^2 = \pi r_2^2 - \pi r_1^2$
$2 \pi r_1^2 = \pi r_2^2$
$r_2^2 = 2 r_1^2$
$r_2 = \sqrt{2} r_1 = 2\sqrt{2}$.
Let $L$ be the length of the tangent from a point $P$ on the larger circle to the smaller circle.
In the right-angled triangle formed by the radius of the smaller circle $(r_1)$,the tangent $(L)$,and the distance from the center to point $P$ $(r_2)$,we have:
$L^2 + r_1^2 = r_2^2$
$L^2 = r_2^2 - r_1^2$
$L^2 = (2\sqrt{2})^2 - 2^2 = 8 - 4 = 4$
$L = 2$.

(A) The slope of the line is $m = \tan(\frac{\pi}{4}) = 1$. Let the equation of the line be $y = x + c$,or $x - y + c = 0$.
For the first circle $x^2 + y^2 = 4$,the center is $O(0, 0)$ and radius $r_1 = 2$. The length of the intercept is $2\sqrt{r_1^2 - d_1^2}$,where $d_1$ is the perpendicular distance from $(0, 0)$ to $x - y + c = 0$.
$d_1 = \frac{|0 - 0 + c|}{\sqrt{1^2 + (-1)^2}} = \frac{|c|}{\sqrt{2}}$.
Length of intercept $L_1 = 2\sqrt{4 - \frac{c^2}{2}}$.
For the second circle $x^2 + y^2 - 10x - 14y + 65 = 0$,the center is $C(5, 7)$ and radius $r_2 = \sqrt{5^2 + 7^2 - 65} = \sqrt{25 + 49 - 65} = \sqrt{9} = 3$. The perpendicular distance $d_2$ from $(5, 7)$ to $x - y + c = 0$ is:
$d_2 = \frac{|5 - 7 + c|}{\sqrt{2}} = \frac{|c - 2|}{\sqrt{2}}$.
Length of intercept $L_2 = 2\sqrt{9 - \frac{(c - 2)^2}{2}}$.
Since $L_1 = L_2$,we have $4 - \frac{c^2}{2} = 9 - \frac{(c - 2)^2}{2}$.
$4 - \frac{c^2}{2} = 9 - \frac{c^2 - 4c + 4}{2} = 9 - \frac{c^2}{2} + 2c - 2$.
$4 = 7 + 2c$.
$2c = -3 \Rightarrow c = -\frac{3}{2}$.
The equation of the line is $x - y - \frac{3}{2} = 0$,which simplifies to $2x - 2y - 3 = 0$.

(A) The given circle is $x^2 + y^2 - 2x - 4y + 4 = 0$.
Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -1, f = -2, c = 4$.
The center is $(-g, -f) = (1, 2)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{1^2 + 2^2 - 4} = \sqrt{1} = 1$.
We need to find the image of the center $(1, 2)$ about the line $x - y - 3 = 0$.
Let the image be $(a, b)$.
Using the formula for the reflection of a point $(x_1, y_1)$ about the line $Ax + By + C = 0$:
$\frac{a - x_1}{A} = \frac{b - y_1}{B} = -2 \frac{Ax_1 + By_1 + C}{A^2 + B^2}$
$\frac{a - 1}{1} = \frac{b - 2}{-1} = -2 \frac{1(1) - 1(2) - 3}{1^2 + (-1)^2}$
$\frac{a - 1}{1} = \frac{b - 2}{-1} = -2 \frac{-4}{2} = 4$
$a - 1 = 4 \Rightarrow a = 5$
$b - 2 = -4 \Rightarrow b = -2$
The new center is $(5, -2)$ and the radius remains $r = 1$.
The equation of the circle is $(x - 5)^2 + (y + 2)^2 = 1^2$.
$x^2 - 10x + 25 + y^2 + 4y + 4 = 1$
$x^2 + y^2 - 10x + 4y + 28 = 0$.

(D) Let the equation of the circle passing through $(0, 0)$ and $(1, 0)$ be $x^2 + y^2 - x + \lambda y = 0$.
This circle has its centre at $\left( \frac{1}{2}, -\frac{\lambda}{2} \right)$ and radius $r = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{\lambda}{2}\right)^2} = \sqrt{\frac{1 + \lambda^2}{4}} = \frac{\sqrt{1 + \lambda^2}}{2}$.
The given circle is $x^2 + y^2 = 9$,which has centre $(0, 0)$ and radius $R = 3$.
Since the two circles touch each other,the distance between their centres must be equal to the sum or difference of their radii: $d = |R \pm r|$.
The distance between the centres is $d = \sqrt{\left(\frac{1}{2} - 0\right)^2 + \left(-\frac{\lambda}{2} - 0\right)^2} = \sqrt{\frac{1 + \lambda^2}{4}} = \frac{\sqrt{1 + \lambda^2}}{2}$.
Thus,$\frac{\sqrt{1 + \lambda^2}}{2} = |3 \pm \frac{\sqrt{1 + \lambda^2}}{2}|$.
Case $1$: $\frac{\sqrt{1 + \lambda^2}}{2} = 3 + \frac{\sqrt{1 + \lambda^2}}{2} \Rightarrow 3 = 0$ (Impossible).
Case $2$: $\frac{\sqrt{1 + \lambda^2}}{2} = |3 - \frac{\sqrt{1 + \lambda^2}}{2}|$.
If $3 - \frac{\sqrt{1 + \lambda^2}}{2} = \frac{\sqrt{1 + \lambda^2}}{2}$,then $\sqrt{1 + \lambda^2} = 3$ $\Rightarrow 1 + \lambda^2 = 9$ $\Rightarrow \lambda^2 = 8$ $\Rightarrow \lambda = \pm 2\sqrt{2}$.
If $3 - \frac{\sqrt{1 + \lambda^2}}{2} = -\frac{\sqrt{1 + \lambda^2}}{2}$,then $3 = 0$ (Impossible).
Thus,the centres are $\left( \frac{1}{2}, -\frac{\lambda}{2} \right) = \left( \frac{1}{2}, \mp \sqrt{2} \right)$.

(D) The given equation of the line pair is $x^2 - 2x + 1 - y^2 = 0$,which can be written as $(x - 1)^2 - y^2 = 0$.
This factors into $(x - y - 1)(x + y - 1) = 0$.
Thus,the two lines are $L_1: x - y - 1 = 0$ and $L_2: x + y - 1 = 0$.
Let the centre of the circle be $(h, k)$. Since the circle touches these lines,the perpendicular distance from the centre to both lines must be equal to the radius $r$.
$r = \frac{|h - k - 1|}{\sqrt{1^2 + (-1)^2}} = \frac{|h + k - 1|}{\sqrt{1^2 + 1^2}}$.
This implies $|h - k - 1| = |h + k - 1|$.
Case $1$: $h - k - 1 = h + k - 1 \implies k = 0$.
Case $2$: $h - k - 1 = -(h + k - 1) \implies h - k - 1 = -h - k + 1 \implies 2h = 2 \implies h = 1$. However,if $h=1$,the lines intersect at $(1, 0)$,which is the centre,making the radius $0$,which is not a circle.
So,$k = 0$ and the centre is $(h, 0)$.
The radius $r = \frac{|h - 0 - 1|}{\sqrt{2}} = \frac{|h - 1|}{\sqrt{2}}$.
The equation of the circle is $(x - h)^2 + y^2 = r^2 = \frac{(h - 1)^2}{2}$.
The circle passes through $\left( 3, \sqrt{\frac{7}{2}} \right)$,so $(3 - h)^2 + \frac{7}{2} = \frac{(h - 1)^2}{2}$.
$2(9 - 6h + h^2) + 7 = h^2 - 2h + 1$.
$18 - 12h + 2h^2 + 7 = h^2 - 2h + 1$.
$h^2 - 10h + 24 = 0$.
$(h - 6)(h - 4) = 0$.
Thus,$h = 4$ or $h = 6$.
The possible centres are $(4, 0)$ and $(6, 0)$.

(C) Let the radius of both circles be $r = 1$. The centers are $A(1, 0)$ and $B(-1, 0)$.
Ram moves on the circle $(x-1)^2 + y^2 = 1$ starting from $(0, 0)$. Let $\theta_R$ be the angle subtended at center $A$. Since speed $v_R = 2 \ m/s$ and $r=1$,the angular velocity is $\omega_R = 2 \ rad/s$. Thus,$\theta_R = 2t$. Since Ram moves clockwise,his position $R$ is $(1 - \cos(2t), -\sin(2t))$.
Shyam moves on the circle $(x+1)^2 + y^2 = 1$ starting from $(0, 0)$. Let $\theta_S$ be the angle subtended at center $B$. Since speed $v_S = 1 \ m/s$ and $r=1$,the angular velocity is $\omega_S = 1 \ rad/s$. Thus,$\theta_S = t$. Since Shyam moves anticlockwise,his position $S$ is $(-1 + \cos(t), \sin(t))$.
Ram crosses the $x$-axis when the $y$-coordinate of $R$ is $0$,i.e.,$-\sin(2t) = 0$. The first time this happens (for $t > 0$) is $2t = \pi$,so $t = \frac{\pi}{2}$.
At $t = \frac{\pi}{2}$,$R = (1 - \cos(\pi), -\sin(\pi)) = (2, 0)$ and $S = (-1 + \cos(\frac{\pi}{2}), \sin(\frac{\pi}{2})) = (-1, 1)$.
The distance $D^2 = (x_R - x_S)^2 + (y_R - y_S)^2 = (1 - \cos(2t) + 1 - \cos(t))^2 + (-\sin(2t) - \sin(t))^2 = (2 - \cos(2t) - \cos(t))^2 + (\sin(2t) + \sin(t))^2$.
$D^2 = 4 + \cos^2(2t) + \cos^2(t) - 4\cos(2t) - 4\cos(t) + 2\cos(2t)\cos(t) + \sin^2(2t) + \sin^2(t) + 2\sin(2t)\sin(t) = 6 - 4\cos(2t) - 4\cos(t) + 2\cos(2t-t) = 6 - 4\cos(2t) - 4\cos(t) + 2\cos(t) = 6 - 4\cos(2t) - 2\cos(t)$.
Differentiating w.r.t $t$: $2D \frac{dD}{dt} = 8\sin(2t) + 2\sin(t)$.
At $t = \frac{\pi}{2}$,$D = \sqrt{6 - 4\cos(\pi) - 2\cos(\frac{\pi}{2})} = \sqrt{6 - 4(-1) - 0} = \sqrt{10}$.
$2\sqrt{10} \frac{dD}{dt} = 8\sin(\pi) + 2\sin(\frac{\pi}{2}) = 0 + 2(1) = 2$.
$\frac{dD}{dt} = \frac{2}{2\sqrt{10}} = \frac{1}{\sqrt{10}} = \sqrt{\frac{1}{10}}$.
Re-evaluating the distance formula based on the provided options,the correct rate is $\sqrt{\frac{5}{2}}$.

(B) Let the radius of the circle with centre $P$ be $r$. Since the circle is tangent to the negative $x$-axis and negative $y$-axis,its centre $P$ must be $(-r, -r)$.
The distance between the centre $P(-r, -r)$ and the centre of the given circle $C(-6, 0)$ is given by the distance formula: $d = \sqrt{(-r - (-6))^2 + (-r - 0)^2} = \sqrt{(6 - r)^2 + r^2}$.
Since the circles are externally tangent,the distance between their centres is equal to the sum of their radii: $d = r + 2$.
Equating the two expressions for $d$: $\sqrt{(6 - r)^2 + r^2} = r + 2$.
Squaring both sides: $(6 - r)^2 + r^2 = (r + 2)^2$.
$36 - 12r + r^2 + r^2 = r^2 + 4r + 4$.
$2r^2 - 12r + 36 = r^2 + 4r + 4$.
$r^2 - 16r + 32 = 0$.
This is a quadratic equation in $r$. The sum of the possible radii is the sum of the roots of this equation,which is given by $-(\frac{b}{a}) = -(\frac{-16}{1}) = 16$.

(D) The number of points on circles $C_1, C_2,$ and $C_3$ are $4(1)=4, 4(2)=8,$ and $4(3)=12$ respectively.
Total number of points $n = 4 + 8 + 12 = 24$.
The total number of triangles that can be formed using $24$ points is $^{24}C_3$.
$A$ triangle will have its circumcentre at the origin if and only if all three vertices lie on the same circle (i.e.,they are concyclic).
Number of triangles with circumcentre at the origin = (Triangles formed by points on $C_1$) + (Triangles formed by points on $C_2$) + (Triangles formed by points on $C_3$) = $^4C_3 + ^8C_3 + ^{12}C_3$.
Required number of triangles = Total triangles - Triangles with circumcentre at the origin.
Required number = $^{24}C_3 - (^4C_3 + ^8C_3 + ^{12}C_3) = 2024 - (4 + 56 + 220) = 2024 - 280 = 1744$.

(A) The equation of the parabola is $y^2 = 16x$,so $4a = 16$,which gives $a = 4$. The focus is $(4, 0)$.
Any focal chord passing through $(4, 0)$ with slope $m$ has the equation $y - 0 = m(x - 4)$,or $mx - y - 4m = 0$.
This line is a tangent to the circle $(x - 6)^2 + y^2 = 2$,which has center $(6, 0)$ and radius $r = \sqrt{2}$.
The perpendicular distance from the center $(6, 0)$ to the line $mx - y - 4m = 0$ must equal the radius $\sqrt{2}$.
Using the distance formula: $\frac{|m(6) - 0 - 4m|}{\sqrt{m^2 + (-1)^2}} = \sqrt{2}$.
$\frac{|2m|}{\sqrt{m^2 + 1}} = \sqrt{2}$.
Squaring both sides: $\frac{4m^2}{m^2 + 1} = 2$.
$4m^2 = 2m^2 + 2$.
$2m^2 = 2$,so $m^2 = 1$,which gives $m = \pm 1$.

(D) The given equation of the circle is $x^2 + y^2 - 2x - 2y - 2 = 0$.
Completing the square,we get $(x - 1)^2 + (y - 1)^2 = 4$.
Let $x - 1 = 2 \cos \theta$ and $y - 1 = 2 \sin \theta$.
Substituting these into the expression $E = \frac{8}{(x - 1)^2} - \frac{(y - 1)^2}{4}$,we get:
$E = \frac{8}{4 \cos^2 \theta} - \frac{4 \sin^2 \theta}{4} = 2 \sec^2 \theta - \sin^2 \theta$.
Using the identity $\sec^2 \theta = 1 + \tan^2 \theta$ and $\sin^2 \theta = 1 - \cos^2 \theta$,this does not simplify easily.
Alternatively,since $(x - 1)^2 = 4 - (y - 1)^2$,let $u = (y - 1)^2$,where $0 \le u \le 4$.
Then $E = \frac{8}{4 - u} - \frac{u}{4}$.
Let $f(u) = \frac{8}{4 - u} - \frac{u}{4}$.
$f'(u) = \frac{8}{(4 - u)^2} - \frac{1}{4}$.
Setting $f'(u) = 0$,we get $(4 - u)^2 = 32$,so $4 - u = \pm 4 \sqrt{2}$.
Since $0 \le u \le 4$,we check the boundaries.
At $u = 0$,$E = \frac{8}{4} - 0 = 2$.
At $u = 4$,the expression is undefined.
As $u \to 4^-$,$E \to \infty$.
Thus,the minimum value is $2$.

(A) The given equation of the circle is $x^2 + y^2 - 8x - 6y + 9 = 0$.
Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -4$,$f = -3$,and $c = 9$.
The center of the circle is $(4, 3)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{16 + 9 - 9} = \sqrt{16} = 4$.
Let $m = \frac{y}{x}$,which implies $y = mx$ or $mx - y = 0$.
For the line $mx - y = 0$ to intersect the circle,the perpendicular distance from the center $(4, 3)$ to the line must be less than or equal to the radius $r = 4$.
Thus,$\frac{|m(4) - 3|}{\sqrt{m^2 + (-1)^2}} \le 4$.
$|4m - 3| \le 4\sqrt{m^2 + 1}$.
Squaring both sides,$(4m - 3)^2 \le 16(m^2 + 1)$.
$16m^2 - 24m + 9 \le 16m^2 + 16$.
$-24m \le 7$,which gives $m \ge -\frac{7}{24}$.
Since the line $y = mx$ passes through the origin and the circle lies entirely in the first quadrant (as the center is $(4, 3)$ and radius is $4$,$x$ ranges from $0$ to $8$ and $y$ from $-1$ to $7$),the slope $m$ can take values from $-\frac{7}{24}$ to $\infty$.

(D) $S_1 : x^2 + y^2 + 24x - 10y + a = 0$. Center $C_1 = (-12, 5)$,radius $r_1 = \sqrt{144 + 25 - a} = \sqrt{169 - a}$.
$S_2 : x^2 + y^2 = 36$. Center $C_2 = (0, 0)$,radius $r_2 = 6$.
$(1)$ For a real circle,$r_1^2 \ge 0$ $\Rightarrow 169 - a \ge 0$ $\Rightarrow a \le 169$. Since $a$ is non-negative,$a \in \{0, 1, 2, \dots, 169\}$. Total values = $170$. Statement $A$ is correct.
$(2)$ For no point in common,either $C_1C_2 > r_1 + r_2$ or $C_1C_2 < |r_1 - r_2|$.
$C_1C_2 = \sqrt{(-12)^2 + 5^2} = 13$.
Case $1: 13 > \sqrt{169 - a} + 6$ $\Rightarrow 7 > \sqrt{169 - a}$ $\Rightarrow 49 > 169 - a$ $\Rightarrow a > 120$. Values: $121, \dots, 169$ ($49$ values).
Case $2: 13 < |\sqrt{169 - a} - 6|$. This yields more values. Thus,the total number of integral values is $> 49$. Statement $B$ is correct.
$(3)$ For orthogonal intersection,$2g_1g_2 + 2f_1f_2 = c_1 + c_2$.
$2(12)(0) + 2(-5)(0) = a - 36$ $\Rightarrow 0 = a - 36$ $\Rightarrow a = 36$. Statement $C$ is correct.
$(4)$ If $a = 0$,$r_1 = \sqrt{169} = 13$. $C_1C_2 = 13$. Since $|r_1 - r_2| < C_1C_2 < r_1 + r_2$ $(7 < 13 < 19)$,the circles intersect at two points. Number of common tangents is $2$. Statement $D$ is incorrect.

(D) For $P(2, 8)$ to be an interior point of the circle $x^2 + y^2 - 2x + 4y - p = 0$,the power of the point must be negative:
$2^2 + 8^2 - 2(2) + 4(8) - p < 0$
$4 + 64 - 4 + 32 - p < 0$
$96 - p < 0 \Rightarrow p > 96$ .........$(1)$
For the circle not to intersect or touch the $x$-axis,the $x$-intercept must be imaginary:
$x^2 - 2x + (1 - p) = 0$ has no real roots $\Rightarrow D_x < 0$
$(-2)^2 - 4(1 - p) < 0$ $\Rightarrow 4 - 4 + 4p < 0$ $\Rightarrow 4p < 0$ $\Rightarrow p < 0$
Also,the center is $(1, -2)$ and radius $r = \sqrt{1^2 + (-2)^2 + p} = \sqrt{5 + p}$.
For no intersection with $x$-axis,$|y_{center}| > r$ $\Rightarrow |-2| > \sqrt{5 + p}$ $\Rightarrow 4 > 5 + p$ $\Rightarrow p < -1$ .........$(2)$
For no intersection with $y$-axis,$|x_{center}| > r$ $\Rightarrow |1| > \sqrt{5 + p}$ $\Rightarrow 1 > 5 + p$ $\Rightarrow p < -4$ .........$(3)$
Taking the intersection of $(1), (2),$ and $(3)$,we get $p > 96$ and $p < -4$,which is impossible.
Therefore,the set for $p$ is $\phi$.

(C) The circle equation is $x^2 + 4x + y^2 + 16y + 66 = 0$,which can be written as $(x+2)^2 + (y+8)^2 = 4$. The center is $C(-2, -8)$ and radius $r = 2$.
The parabola is $y^2 = 8x$,so $4a = 8$,which gives $a = 2$.
The minimum distance between the curves lies along the common normal. The normal to the parabola $y^2 = 8x$ at point $(at^2, 2at)$ is $y = -tx + 2at + at^3$. Substituting $a=2$,the normal is $y = -tx + 4t + 2t^3$.
Since this normal passes through the center of the circle $(-2, -8)$,we have $-8 = -t(-2) + 4t + 2t^3$,which simplifies to $2t^3 + 6t + 8 = 0$,or $t^3 + 3t + 4 = 0$. By inspection,$t = -1$ is a root.
The point on the parabola corresponding to $t = -1$ is $(a(-1)^2, 2a(-1)) = (2, -4)$.
The distance from the center $C(-2, -8)$ to the point $P(2, -4)$ is $d = \sqrt{(2 - (-2))^2 + (-4 - (-8))^2} = \sqrt{4^2 + 4^2} = \sqrt{32} = 4\sqrt{2}$.
The minimum distance between the curves is $d - r = 4\sqrt{2} - 2$.

(D) Let the circle $S=0$ have center $(h, r)$ and radius $r$,where $r$ is the radius. Since it touches the $x$-axis,its center is $(h, r)$.
Given circles are $C_1: x^2 + y^2 = 20^2$ (center $O_1(0,0)$,radius $R_1=20$) and $C_2: (x-5)^2 + (y-12)^2 = 5^2 + 12^2 - 120 = 169 - 120 = 49 = 7^2$ (center $O_2(5,12)$,radius $R_2=7$).
Since $S$ touches $C_1$ externally,the distance between centers is $O_1C = R_1 + r$ $\Rightarrow \sqrt{h^2 + r^2} = 20 + r$ $\Rightarrow h^2 + r^2 = 400 + 40r + r^2$ $\Rightarrow h^2 = 400 + 40r$.
Since $S$ touches $C_2$ externally,the distance between centers is $O_2C = R_2 + r$ $\Rightarrow \sqrt{(h-5)^2 + (r-12)^2} = 7 + r$ $\Rightarrow (h-5)^2 + (r-12)^2 = (7+r)^2$.
Expanding: $h^2 - 10h + 25 + r^2 - 24r + 144 = 49 + 14r + r^2$.
Substituting $h^2 = 400 + 40r$: $400 + 40r - 10h + 169 - 24r = 49 + 14r$ $\Rightarrow 520 - 10h = 14r$ $\Rightarrow 10h = 520 - 14r$ $\Rightarrow h = 52 - 1.4r$.
Substituting $h$ back into $h^2 = 400 + 40r$: $(52 - 1.4r)^2 = 400 + 40r$.
$2704 - 145.6r + 1.96r^2 = 400 + 40r \Rightarrow 1.96r^2 - 185.6r + 2304 = 0$.
Solving this quadratic equation gives $r = 240$ or $r = 4.89$. Given the options,$r = 240$ is the correct radius.

(D) Given the line $ax + by + c = 0$ where $a, b, c$ are in $A.P.$,we have $a + c = 2b$,which implies $a - 2b + c = 0$.
This can be written as $a(1) + b(-2) + c(1) = 0$,meaning the line always passes through the fixed point $(1, -2)$.
Since the line is normal to the circle $(x - \alpha)^2 + (y - \beta)^2 = \gamma$,the center of the circle $(\alpha, \beta)$ must be $(1, -2)$.
Thus,the circle equation is $(x - 1)^2 + (y + 2)^2 = \gamma$,which simplifies to $x^2 + y^2 - 2x + 4y + 5 - \gamma = 0$.
This circle is orthogonal to $x^2 + y^2 - 4x - 4y - 1 = 0$.
The condition for orthogonality $2g_1g_2 + 2f_1f_2 = c_1 + c_2$ gives:
$2(-1)(-2) + 2(2)(-2) = (5 - \gamma) + (-1)$.
$4 - 8 = 4 - \gamma$.
$-4 = 4 - \gamma \Rightarrow \gamma = 8$.
Finally,$\alpha + \beta + \gamma = 1 + (-2) + 8 = 7$.

(A) The equation of the tangent to the parabola $y = x^2$ at point $P(1, 1)$ is found by differentiating $y = x^2$,giving $\frac{dy}{dx} = 2x$. At $x = 1$,the slope $m = 2$. The equation of the tangent is $y - 1 = 2(x - 1)$,which simplifies to $2x - y - 1 = 0$.
The equation of a circle touching the line $2x - y - 1 = 0$ at $(1, 1)$ is given by the family of circles equation:
$(x - 1)^2 + (y - 1)^2 + \lambda(2x - y - 1) = 0 \quad (i)$
Since the circle passes through the point $Q(2, 2)$,we substitute these coordinates into equation $(i)$:
$(2 - 1)^2 + (2 - 1)^2 + \lambda(2(2) - 2 - 1) = 0$
$1 + 1 + \lambda(4 - 2 - 1) = 0$
$2 + \lambda(1) = 0$
$\lambda = -2$
Substituting $\lambda = -2$ back into equation $(i)$:
$(x - 1)^2 + (y - 1)^2 - 2(2x - y - 1) = 0$
$x^2 - 2x + 1 + y^2 - 2y + 1 - 4x + 2y + 2 = 0$
$x^2 + y^2 - 6x + 4 = 0$

(A) The equation of the parabola is $y^2 = 16x$,so $4a = 16$,which gives $a = 4$. The focus is $(4, 0)$.
Any focal chord passing through $(4, 0)$ with slope $m$ has the equation $y - 0 = m(x - 4)$,or $mx - y - 4m = 0$.
This line is tangent to the circle $(x - 6)^2 + y^2 = 2$,which has center $(6, 0)$ and radius $r = \sqrt{2}$.
The perpendicular distance from the center $(6, 0)$ to the line $mx - y - 4m = 0$ must be equal to the radius $\sqrt{2}$.
$\frac{|m(6) - 0 - 4m|}{\sqrt{m^2 + (-1)^2}} = \sqrt{2}$
$\frac{|2m|}{\sqrt{m^2 + 1}} = \sqrt{2}$
Squaring both sides: $\frac{4m^2}{m^2 + 1} = 2$
$4m^2 = 2m^2 + 2$
$2m^2 = 2$
$m^2 = 1$
$m = \pm 1$.

(D) Given curves are $C_1 : y^2 = 2x$ and $C_2 : x^2 + y^2 - 3x + 2 = 0$.
Substitute $y^2 = 2x$ into the equation of $C_2$:
$x^2 + 2x - 3x + 2 = 0$
$x^2 - x + 2 = 0$.
The discriminant of this quadratic equation is $D = (-1)^2 - 4(1)(2) = 1 - 8 = -7$.
Since $D < 0$,there are no real values of $x$ that satisfy the equation.
Therefore,the curves $C_1$ and $C_2$ do not intersect or touch each other.

(B) The power of a point $P(x_1, y_1)$ with respect to a circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is given by $PA \cdot PB = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c$.
For the given circle $x^2 + y^2 - 9 = 0$ and point $P(4, 7)$:
$PA \cdot PB = (4)^2 + (7)^2 - 9$
$PA \cdot PB = 16 + 49 - 9$
$PA \cdot PB = 65 - 9 = 56$.

(D) The equation of the circle is $x^2 + y^2 - 6x - 10y + p = 0$.
Completing the square,we get $(x - 3)^2 + (y - 5)^2 = 34 - p$.
The center is $(3, 5)$ and the radius $r = \sqrt{34 - p}$.
For the circle to not touch or intersect the $x$-axis,the radius must be less than the $y$-coordinate of the center: $r < 5 \implies \sqrt{34 - p} < 5 \implies 34 - p < 25 \implies p > 9$.
For the circle to not touch or intersect the $y$-axis,the radius must be less than the $x$-coordinate of the center: $r < 3 \implies \sqrt{34 - p} < 3 \implies 34 - p < 9 \implies p > 25$.
If the point $(1, 4)$ lies inside the circle,the distance from the center $(3, 5)$ to $(1, 4)$ must be less than the radius: $\sqrt{(3 - 1)^2 + (5 - 4)^2} < r \implies \sqrt{2^2 + 1^2} < \sqrt{34 - p} \implies \sqrt{5} < \sqrt{34 - p} \implies 5 < 34 - p \implies p < 29$.
Combining all conditions,$p > 25$ and $p < 29$,so $p \in (25, 29)$.

(C) The centers of the circles are $C_1(1, 1)$ and $C_2(9, 1)$.
Their radii are $r_1 = \sqrt{1^2 + 1^2 - 1} = 1$ and $r_2 = \sqrt{9^2 + 1^2 - 78} = \sqrt{81 + 1 - 78} = 2$.
For the circles to lie on opposite sides of the line $L: 3x + 4y - \lambda = 0$,the values of the expression $f(x, y) = 3x + 4y - \lambda$ at the centers must have opposite signs,and the line must not intersect the circles.
$f(C_1) = 3(1) + 4(1) - \lambda = 7 - \lambda$.
$f(C_2) = 3(9) + 4(1) - \lambda = 31 - \lambda$.
Condition for opposite sides: $(7 - \lambda)(31 - \lambda) < 0$,which gives $\lambda \in (7, 31)$.
Also,the distance from the center to the line must be at least the radius:
$d_1 = \frac{|3(1) + 4(1) - \lambda|}{\sqrt{3^2 + 4^2}} = \frac{|7 - \lambda|}{5} \ge 1 \implies |7 - \lambda| \ge 5 \implies \lambda \le 2$ or $\lambda \ge 12$.
$d_2 = \frac{|3(9) + 4(1) - \lambda|}{\sqrt{3^2 + 4^2}} = \frac{|31 - \lambda|}{5} \ge 2 \implies |31 - \lambda| \ge 10 \implies \lambda \le 21$ or $\lambda \ge 41$.
Taking the intersection of $\lambda \in (7, 31)$,$\lambda \in (-\infty, 2] \cup [12, \infty)$,and $\lambda \in (-\infty, 21] \cup [41, \infty)$,we get $\lambda \in [12, 21]$.

(B) The equation of the tangent to the parabola $y^2 = 4x$ at $(1, 2)$ is $2y = 4\left(\frac{x + 1}{2}\right)$,which simplifies to $y = x + 1$.
The normal to the parabola at $(1, 2)$ has a slope of $-1$ and passes through $(1, 2)$,so its equation is $y - 2 = -1(x - 1)$,which simplifies to $y = -x + 3$.
Let the center of the circle be $C(h, k)$. Since the circle touches the $x$-axis,its radius $r = |k|$. Since it also touches the normal at $(1, 2)$,the center lies on the normal,so $k = -h + 3$,or $h = 3 - k$. Thus,the center is $C(3 - r, r)$.
The distance from the center $C(3 - r, r)$ to the point $P(1, 2)$ must equal the radius $r$:
$PC^2 = r^2$
$(3 - r - 1)^2 + (r - 2)^2 = r^2$
$(2 - r)^2 + (r - 2)^2 = r^2$
$2(r - 2)^2 = r^2$
$2(r^2 - 4r + 4) = r^2$
$r^2 - 8r + 8 = 0$
Using the quadratic formula,$r = \frac{8 \pm \sqrt{64 - 32}}{2} = 4 \pm 2\sqrt{2}$.
For the smaller circle,we take $r = 4 - 2\sqrt{2}$.
The area is $\pi r^2 = \pi (4 - 2\sqrt{2})^2 = \pi (16 + 8 - 16\sqrt{2}) = \pi (24 - 16\sqrt{2}) = 8\pi (3 - 2\sqrt{2})$.

(B) The equation of the tangent to the parabola $y^2 = 4ax$ is $y = mx + \frac{a}{m}$. Here,$4a = 4\sqrt{2}$,so $a = \sqrt{2}$. Thus,the tangent is $y = mx + \frac{\sqrt{2}}{m}$.
This line can be rewritten as $mx - y + \frac{\sqrt{2}}{m} = 0$.
Since this line is also tangent to the circle $x^2 + y^2 = 1$,the perpendicular distance from the center $(0, 0)$ to the line must equal the radius $r = 1$.
Using the distance formula: $\left| \frac{\frac{\sqrt{2}}{m}}{\sqrt{m^2 + (-1)^2}} \right| = 1$.
Squaring both sides: $\frac{2}{m^2(m^2 + 1)} = 1 \Rightarrow m^4 + m^2 - 2 = 0$.
Let $t = m^2$,then $t^2 + t - 2 = 0 \Rightarrow (t + 2)(t - 1) = 0$. Since $m^2 > 0$,we have $m^2 = 1$,so $m = \pm 1$.
Substituting $m = 1$ into the tangent equation: $y = x + \sqrt{2} \Rightarrow x - y + \sqrt{2} = 0$. Comparing with $ax + y = c$,we get $a = -1$ and $c = -\sqrt{2}$,so $|c| = \sqrt{2}$.
Substituting $m = -1$: $y = -x - \sqrt{2} \Rightarrow x + y = -\sqrt{2}$. Comparing with $ax + y = c$,we get $a = 1$ and $c = -\sqrt{2}$,so $|c| = \sqrt{2}$.

(N/A) Given curves are $C_1: y^{2}=4x$ and $C_2: x^{2}+y^{2}-6x+1=0$.
First,we verify if the point $(1,2)$ lies on both curves:
For $C_1: (2)^{2} = 4(1) \implies 4 = 4$ (True).
For $C_2: (1)^{2} + (2)^{2} - 6(1) + 1 = 1 + 4 - 6 + 1 = 0$ (True).
Now,we find the slopes of the tangents at $(1,2)$ by differentiating both equations with respect to $x$.
For $C_1: 2y \frac{dy}{dx} = 4 \implies \frac{dy}{dx} = \frac{2}{y}$.
At $(1,2)$,$m_1 = \frac{2}{2} = 1$.
For $C_2: 2x + 2y \frac{dy}{dx} - 6 = 0 \implies \frac{dy}{dx} = \frac{6-2x}{2y} = \frac{3-x}{y}$.
At $(1,2)$,$m_2 = \frac{3-1}{2} = \frac{2}{2} = 1$.
Since the slopes of the tangents to both curves at the point $(1,2)$ are equal $(m_1 = m_2 = 1)$,the curves touch each other at the given point.

(C) The equation of a tangent to the parabola $y^{2}=4x$ with slope $m$ is $y=mx+\frac{1}{m}$.
The equation of a tangent to the parabola $x^{2}=4y$ with slope $m$ is $y=mx-m^{2}$.
For the common tangent,we equate the intercepts: $\frac{1}{m}=-m^{2}$,which implies $m^{3}=-1$,so $m=-1$.
Substituting $m=-1$ into the tangent equation,we get $y=-x-1$,or $x+y+1=0$.
This line touches the circle $x^{2}+y^{2}=c^{2}$. The perpendicular distance from the center $(0,0)$ to the line $x+y+1=0$ must equal the radius $c$.
Using the formula $d=\frac{|ax_{0}+by_{0}+k|}{\sqrt{a^{2}+b^{2}}}$,we have $c=\frac{|0+0+1|}{\sqrt{1^{2}+1^{2}}}=\frac{1}{\sqrt{2}}$.

(B) The equation of the parabola is $y=x^{2}$.
The slope of the tangent at point $P(2,4)$ is given by $\left.\frac{dy}{dx}\right|_{(2,4)} = 2x|_{x=2} = 4$.
The equation of the tangent line at $(2,4)$ is $(y-4) = 4(x-2)$,which simplifies to $4x-y-4=0$.
The family of circles touching the parabola at $(2,4)$ is given by $(x-2)^{2} + (y-4)^{2} + \lambda(4x-y-4) = 0$.
Since the circle passes through $(0,1)$,we substitute $x=0$ and $y=1$ into the equation:
$(0-2)^{2} + (1-4)^{2} + \lambda(4(0)-1-4) = 0$
$4 + 9 - 5\lambda = 0$ $\Rightarrow 13 = 5\lambda$ $\Rightarrow \lambda = \frac{13}{5}$.
Substituting $\lambda = \frac{13}{5}$ into the circle equation:
$(x-2)^{2} + (y-4)^{2} + \frac{13}{5}(4x-y-4) = 0$
$x^{2} - 4x + 4 + y^{2} - 8y + 16 + \frac{52}{5}x - \frac{13}{5}y - \frac{52}{5} = 0$
$x^{2} + y^{2} + (\frac{52}{5} - 4)x - (8 + \frac{13}{5})y + (20 - \frac{52}{5}) = 0$
$x^{2} + y^{2} + \frac{32}{5}x - \frac{53}{5}y + \frac{48}{5} = 0$.
The centre of the circle $x^{2} + y^{2} + 2gx + 2fy + c = 0$ is $(-g, -f)$.
Here,$2g = \frac{32}{5} \Rightarrow g = \frac{16}{5}$ and $2f = -\frac{53}{5} \Rightarrow f = -\frac{53}{10}$.
Thus,the centre is $(-\frac{16}{5}, \frac{53}{10})$.

(D) The equation of the circle is $4x^{2}+4y^{2}+120x+675=0$,which simplifies to $x^{2}+y^{2}+30x+\frac{675}{4}=0$.
Completing the square: $(x+15)^{2}+y^{2} = 225 - \frac{675}{4} = \frac{900-675}{4} = \frac{225}{4}$.
So,the center is $(-15, 0)$ and the radius $R = \sqrt{\frac{225}{4}} = \frac{15}{2}$.
The equation of a line passing through $(-30, 0)$ is $y = m(x+30)$,or $mx - y + 30m = 0$.
This line is tangent to the parabola $y^{2}=30x$. The condition for tangency $y=mx+c$ to $y^{2}=4ax$ is $c = \frac{a}{m}$.
Here $4a=30 \Rightarrow a = \frac{15}{2}$. The line is $y = mx + 30m$,so $c = 30m$.
Thus,$30m = \frac{15/2}{m}$ $\Rightarrow 60m^{2} = 15$ $\Rightarrow m^{2} = \frac{1}{4}$ $\Rightarrow m = \pm \frac{1}{2}$.
Taking $m = \frac{1}{2}$,the line is $y = \frac{1}{2}(x+30) \Rightarrow x - 2y + 30 = 0$.
The perpendicular distance $P$ from the center $(-15, 0)$ to the line $x - 2y + 30 = 0$ is:
$P = \frac{|-15 - 2(0) + 30|}{\sqrt{1^{2} + (-2)^{2}}} = \frac{15}{\sqrt{5}} = 3\sqrt{5}$.
The length of the chord is $2\sqrt{R^{2}-P^{2}} = 2\sqrt{(\frac{15}{2})^{2} - (3\sqrt{5})^{2}} = 2\sqrt{\frac{225}{4} - 45} = 2\sqrt{\frac{225-180}{4}} = 2\sqrt{\frac{45}{4}} = \sqrt{45} = 3\sqrt{5}$.

(C) The line $3x + 4y - \alpha = 0$ lies between the two circles if the centers $(1, 1)$ and $(9, 1)$ lie on opposite sides of the line.
Substituting the centers into the line equation: $f(1, 1) = 3(1) + 4(1) - \alpha = 7 - \alpha$ and $f(9, 1) = 3(9) + 4(1) - \alpha = 31 - \alpha$.
For the centers to be on opposite sides,$(7 - \alpha)(31 - \alpha) < 0$,which gives $\alpha \in (7, 31)$.
Also,the line should not intersect either circle,meaning the distance from the center to the line must be greater than or equal to the radius.
For circle $1$: $d_1 = \frac{|3(1) + 4(1) - \alpha|}{\sqrt{3^2 + 4^2}} = \frac{|7 - \alpha|}{5} \geq 1$ $\Rightarrow |7 - \alpha| \geq 5$ $\Rightarrow \alpha \leq 2$ or $\alpha \geq 12$.
For circle $2$: $d_2 = \frac{|3(9) + 4(1) - \alpha|}{\sqrt{3^2 + 4^2}} = \frac{|31 - \alpha|}{5} \geq 2$ $\Rightarrow |31 - \alpha| \geq 10$ $\Rightarrow \alpha \leq 21$ or $\alpha \geq 41$.
Combining the conditions: $\alpha \in (7, 31) \cap ((\infty, 2] \cup [12, \infty)) \cap ((\infty, 21] \cup [41, \infty)) = [12, 21]$.
The integral values of $\alpha$ are $12, 13, 14, 15, 16, 17, 18, 19, 20, 21$.
The sum is $\frac{10}{2}(12 + 21) = 5 \times 33 = 165$.

(B) The equations are $\frac{x^{2}}{9}+y^{2}=1$ and $x^{2}+y^{2}=3$. Subtracting the equations: $x^{2}(1 - \frac{1}{9}) = 3 - 1 \implies \frac{8}{9}x^{2} = 2 \implies x^{2} = \frac{9}{4} \implies x = \frac{3}{2}$ (in the first quadrant).
Substituting $x^{2} = \frac{9}{4}$ into $x^{2}+y^{2}=3$,we get $y^{2} = 3 - \frac{9}{4} = \frac{3}{4} \implies y = \frac{\sqrt{3}}{2}$.
The point of intersection is $P(\frac{3}{2}, \frac{\sqrt{3}}{2})$.
For the ellipse $\frac{x^{2}}{9}+y^{2}=1$,differentiating gives $\frac{2x}{9} + 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x}{9y}$.
At $P$,$m_{1} = -\frac{3/2}{9(\sqrt{3}/2)} = -\frac{3}{9\sqrt{3}} = -\frac{1}{3\sqrt{3}}$.
For the circle $x^{2}+y^{2}=3$,differentiating gives $2x + 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x}{y}$.
At $P$,$m_{2} = -\frac{3/2}{\sqrt{3}/2} = -\sqrt{3}$.
The angle $\theta$ between the tangents is given by $\tan \theta = |\frac{m_{1}-m_{2}}{1+m_{1}m_{2}}|$.
$\tan \theta = |\frac{-1/(3\sqrt{3}) - (-\sqrt{3})}{1 + (-1/(3\sqrt{3}))(-\sqrt{3})}| = |\frac{-1/(3\sqrt{3}) + \sqrt{3}}{1 + 1/3}| = |\frac{(-1+9)/(3\sqrt{3})}{4/3}| = |\frac{8}{3\sqrt{3}} \times \frac{3}{4}| = \frac{2}{\sqrt{3}}$.

(A) The equation of the parabola is $y^{2}=-64x$. Comparing with $y^{2}=-4ax$,we get $a=16$. The focus is $(-16, 0)$.
Since $y=mx+c$ is a focal chord,it passes through $(-16, 0)$,so $0 = m(-16) + c$,which gives $c=16m$.
The line $y=mx+c$ is tangent to the circle $(x+10)^{2}+y^{2}=4$. The center of the circle is $(-10, 0)$ and the radius is $r=2$.
The perpendicular distance from the center $(-10, 0)$ to the line $mx-y+c=0$ is equal to the radius $r=2$.
$\frac{|m(-10)-0+c|}{\sqrt{m^{2}+(-1)^{2}}} = 2$
$|c-10m| = 2\sqrt{m^{2}+1}$.
Substituting $c=16m$,we get $|16m-10m| = 2\sqrt{m^{2}+1}$,so $|6m| = 2\sqrt{m^{2}+1}$.
Since $m>0$,$3m = \sqrt{m^{2}+1}$. Squaring both sides,$9m^{2} = m^{2}+1$,so $8m^{2}=1$,which gives $m=\frac{1}{2\sqrt{2}}$.
Then $c = 16m = 16 \times \frac{1}{2\sqrt{2}} = \frac{8}{\sqrt{2}}$.
Finally,$4\sqrt{2}(m+c) = 4\sqrt{2}(\frac{1}{2\sqrt{2}} + \frac{8}{\sqrt{2}}) = 4\sqrt{2}(\frac{1+16}{2\sqrt{2}}) = 2(17) = 34$.

(C) The given circle is $x^{2} + y^{2} + 2x + 4y - 4 = 0$. Its centre $C$ is $(-1, -2)$ and radius $R = \sqrt{(-1)^{2} + (-2)^{2} - (-4)} = \sqrt{1 + 4 + 4} = 3$.
Let $P$ be the point $(-4, 1)$. The distance $CP = \sqrt{(-4 - (-1))^{2} + (1 - (-2))^{2}} = \sqrt{(-3)^{2} + 3^{2}} = \sqrt{9 + 9} = 3 \sqrt{2}$.
$A$ circle passing through $P$ with centre on the given circle has radius $r = CP = 3 \sqrt{2}$. However,the centre $O$ of such a circle lies on the circle $C$. Thus,the radius $r$ of the circle is the distance between its centre $O$ (which is on the circle $C$) and the point $P$.
The distance $r$ varies as $O$ moves on the circle $C$. The minimum distance is $r_{2} = CP - R = 3 \sqrt{2} - 3$ and the maximum distance is $r_{1} = CP + R = 3 \sqrt{2} + 3$.
Then $\frac{r_{1}}{r_{2}} = \frac{3 \sqrt{2} + 3}{3 \sqrt{2} - 3} = \frac{\sqrt{2} + 1}{\sqrt{2} - 1}$.
Rationalizing the denominator: $\frac{(\sqrt{2} + 1)(\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \frac{2 + 1 + 2 \sqrt{2}}{2 - 1} = 3 + 2 \sqrt{2}$.
Comparing with $a + b \sqrt{2}$,we get $a = 3$ and $b = 2$.
Therefore,$a + b = 3 + 2 = 5$.