Geometrical problems regarding circle and its properties Questions in English

(D) The length of the intercept made by the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ on the $x$-axis is $2\sqrt{g^2 - c}$.
Since the circle passes through the origin $(c=0)$,the intercept on the $x$-axis is $2\sqrt{g^2} = 2|g|$.
Given $2|g| = 10$,we get $|g| = 5$.
Similarly,the intercept on the $y$-axis is $2\sqrt{f^2} = 2|f|$.
Given $2|f| = 24$,we get $|f| = 12$.
The radius of the circle is $r = \sqrt{g^2 + f^2 - c}$.
Since $c = 0$,$r = \sqrt{g^2 + f^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$.

(D) The first circle $x^2 + y^2 = a^2$ has center $C_1 = (0, 0)$ and radius $r_1 = a$.
The second circle $x^2 + y^2 - 2gx + g^2 - b^2 = 0$ can be rewritten as $(x - g)^2 + y^2 = b^2$,which has center $C_2 = (g, 0)$ and radius $r_2 = b$.
Two circles touch each other externally if the distance between their centers is equal to the sum of their radii,i.e.,$d(C_1, C_2) = r_1 + r_2$.
The distance between $C_1(0, 0)$ and $C_2(g, 0)$ is $\sqrt{(g - 0)^2 + (0 - 0)^2} = |g|$.
Thus,$|g| = a + b$. Assuming $g, a, b > 0$,we get $g = a + b$.

(A) By the geometric property of circles,exactly one unique circle can pass through three non-collinear points. This circle is known as the circumcircle of the triangle formed by these three points.

(A) For the circle $x^2 + y^2 - 4y = 0$,the center $C_1 = (0, 2)$ and radius $r_1 = \sqrt{0^2 + 2^2 - 0} = 2$.
For the circle $x^2 + y^2 - 8y = 0$,the center $C_2 = (0, 4)$ and radius $r_2 = \sqrt{0^2 + 4^2 - 0} = 4$.
The distance between the centers $C_1$ and $C_2$ is $d = \sqrt{(0-0)^2 + (4-2)^2} = 2$.
Since $|r_2 - r_1| = |4 - 2| = 2$,we observe that $d = |r_2 - r_1|$.
Therefore,the circles touch each other internally.

(D) The first circle is $x^2 + y^2 = 3^2$,with center $C_1 = (0, 0)$ and radius $r_1 = 3$.
The second circle is $x^2 + y^2 + 2ax + 2y + 1 = 0$,with center $C_2 = (-a, -1)$ and radius $r_2 = \sqrt{(-a)^2 + (-1)^2 - 1} = \sqrt{a^2}$. Since $r_2$ must be positive,$r_2 = |a|$.
The distance between centers is $d = \sqrt{(-a - 0)^2 + (-1 - 0)^2} = \sqrt{a^2 + 1}$.
Case $1$: Circles touch externally,$d = r_1 + r_2$.
$\sqrt{a^2 + 1} = 3 + |a|$. Squaring both sides: $a^2 + 1 = 9 + a^2 + 6|a| \Rightarrow 6|a| = -8$,which has no real solution.
Case $2$: Circles touch internally,$d = |r_1 - r_2|$.
$\sqrt{a^2 + 1} = |3 - |a||$. Squaring both sides: $a^2 + 1 = 9 + a^2 - 6|a|$ $\Rightarrow 6|a| = 8$ $\Rightarrow |a| = 4/3$.
Thus,$a = 4/3$ or $a = -4/3$.

(B) The given circle is $x(x - 4) + y(y - 3) = 0$,which simplifies to $x^2 + y^2 - 4x - 3y = 0$.
Its center $C_1$ is $(2, 3/2)$ and radius $r_1 = \sqrt{2^2 + (3/2)^2} = \sqrt{4 + 9/4} = \sqrt{25/4} = 5/2$.
The normals are given by $x^2 + 2xy + 3x + 6y = 0$,which factors as $x(x + 2y) + 3(x + 2y) = 0$,or $(x + 3)(x + 2y) = 0$.
The intersection of these lines gives the center $C_2$ of the required circle. Solving $x + 3 = 0$ and $x + 2y = 0$,we get $x = -3$ and $y = 3/2$. So $C_2 = (-3, 3/2)$.
The distance between centers $C_1(2, 3/2)$ and $C_2(-3, 3/2)$ is $d = \sqrt{(2 - (-3))^2 + (3/2 - 3/2)^2} = 5$.
Since the circle $C_2$ contains $C_1$,the condition is $d = r_2 - r_1$,where $r_2$ is the radius of the required circle.
$5 = r_2 - 5/2 \Rightarrow r_2 = 5 + 5/2 = 15/2$.
The equation of the circle is $(x + 3)^2 + (y - 3/2)^2 = (15/2)^2$.
$x^2 + 6x + 9 + y^2 - 3y + 9/4 = 225/4$.
$x^2 + y^2 + 6x - 3y + 9 + 9/4 - 225/4 = 0$.
$x^2 + y^2 + 6x - 3y + 9 - 216/4 = 0$.
$x^2 + y^2 + 6x - 3y + 9 - 54 = 0$.
$x^2 + y^2 + 6x - 3y - 45 = 0$.

(B) The given circles are ${C_1}: x^2 + (y - 1)^2 = 3^2$ and ${C_2}: (x - 1)^2 + y^2 = 5^2$.
The centers and radii are:
${C_1} = (0, 1)$ with radius ${r_1} = 3$.
${C_2} = (1, 0)$ with radius ${r_2} = 5$.
The distance between the centers is $d = \sqrt{(1 - 0)^2 + (0 - 1)^2} = \sqrt{1^2 + (-1)^2} = \sqrt{2}$.
We compare $d$ with the difference of the radii $|r_2 - r_1| = |5 - 3| = 2$.
Since $d = \sqrt{2} \approx 1.414$ and $|r_2 - r_1| = 2$,we have $d < |r_2 - r_1|$.
When the distance between centers is less than the difference of the radii,one circle lies entirely inside the other.

(B) The given equations of the circles are:
$(x + a)^2 + (y + b)^2 = a^2 \implies x^2 + y^2 + 2ax + 2by + b^2 = 0$
$(x + \alpha)^2 + (y + \beta)^2 = \beta^2 \implies x^2 + y^2 + 2\alpha x + 2\beta y + \alpha^2 = 0$
Comparing these with the general form $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$,we get:
$g_1 = a, f_1 = b, c_1 = b^2$
$g_2 = \alpha, f_2 = \beta, c_2 = \alpha^2$
Two circles cut orthogonally if $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.
Substituting the values:
$2(a\alpha) + 2(b\beta) = b^2 + \alpha^2$
$2(a\alpha + b\beta) = b^2 + \alpha^2$.

(D) The equation of the circle is $x^2 + y^2 - 20y + 90 = 0$.
Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = 0$,$f = -10$,and $c = 90$.
The center of the circle is $(-g, -f) = (0, 10)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{0^2 + (-10)^2 - 90} = \sqrt{100 - 90} = \sqrt{10}$.
For the line $mx - y = 0$ to be outside the circle,the perpendicular distance from the center $(0, 10)$ to the line must be greater than the radius $r$.
Distance $d = \frac{|m(0) - 1(10)|}{\sqrt{m^2 + (-1)^2}} = \frac{10}{\sqrt{m^2 + 1}}$.
Setting $d > r$,we have $\frac{10}{\sqrt{m^2 + 1}} > \sqrt{10}$.
Squaring both sides,$\frac{100}{m^2 + 1} > 10$.
$10 > m^2 + 1$,which implies $m^2 < 9$.
Thus,$|m| < 3$.

(B) The radical axis of two circles is defined as the locus of a point from which the lengths of the tangents drawn to the two circles are equal.
Mathematically,for two circles $S_1 = 0$ and $S_2 = 0$,the radical axis is given by $S_1 - S_2 = 0$.
It is a fundamental property of geometry that the radical axis of two circles is always perpendicular to the line joining their centres.

(C) Given equations of the circles:
$x^2 + y^2 - 2x + 6y + 6 = 0$ $(i)$
$x^2 + y^2 - 5x + 6y + 15 = 0$ $(ii)$
Comparing with the standard equation $x^2 + y^2 + 2gx + 2fy + c = 0$:
For circle $(i)$: $g = -1, f = 3, c = 6$. Centre $A = (1, -3)$,radius $r_1 = \sqrt{(-1)^2 + 3^2 - 6} = \sqrt{1 + 9 - 6} = 2$.
For circle $(ii)$: $g = -2.5, f = 3, c = 15$. Centre $B = (2.5, -3)$,radius $r_2 = \sqrt{(-2.5)^2 + 3^2 - 15} = \sqrt{6.25 + 9 - 15} = \sqrt{0.25} = 0.5$.
Distance between centres $A$ and $B$ is $d = \sqrt{(2.5 - 1)^2 + (-3 - (-3))^2} = \sqrt{1.5^2 + 0^2} = 1.5$.
Difference of radii is $|r_1 - r_2| = |2 - 0.5| = 1.5$.
Since $d = |r_1 - r_2|$,the circles touch each other internally.

(C) The angle subtended by a chord at the center is twice the angle subtended by it at the major segment.
Given the angle at the major segment is ${45^\circ}$,the angle at the center $C(0,0)$ is $2 \times {45^\circ} = {90^\circ}$.
Let $P$ be the foot of the perpendicular from the center $C(0,0)$ to the chord $y = mx + 1$.
In the right-angled triangle $CPR$,where $CR$ is the radius $r = 1$,the angle $\angle PCR = {45^\circ}$.
Thus,$CP = r \cos({45^\circ}) = 1 \times \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}$.
The perpendicular distance from $(0,0)$ to $mx - y + 1 = 0$ is given by $\frac{|m(0) - 0 + 1|}{\sqrt{m^2 + (-1)^2}} = \frac{1}{\sqrt{m^2 + 1}}$.
Equating the two expressions for $CP$:
$\frac{1}{\sqrt{m^2 + 1}} = \frac{1}{\sqrt{2}}$
$m^2 + 1 = 2$
$m^2 = 1$
$m = \pm 1$.
Since the options provided include $-1$ but not $1$,and the question asks for the value of $m$,the correct choice is $-1$.

(A) The first circle is $(x - 1)^2 + (y - 3)^2 = r^2$,with center $C_1 = (1, 3)$ and radius $r_1 = r$.
The second circle is $x^2 + y^2 - 8x + 2y + 8 = 0$. Rewriting in standard form: $(x^2 - 8x + 16) + (y^2 + 2y + 1) = -8 + 16 + 1$,which is $(x - 4)^2 + (y + 1)^2 = 9 = 3^2$. Thus,center $C_2 = (4, -1)$ and radius $r_2 = 3$.
The distance between the centers $d = \sqrt{(4 - 1)^2 + (-1 - 3)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = 5$.
For two circles to intersect at two distinct points,the condition is $|r_1 - r_2| < d < r_1 + r_2$.
$1$) $r_1 + r_2 > d$ $\Rightarrow r + 3 > 5$ $\Rightarrow r > 2$.
$2$) $|r_1 - r_2| < d$ $\Rightarrow |r - 3| < 5$ $\Rightarrow -5 < r - 3 < 5$ $\Rightarrow -2 < r < 8$. Since $r$ must be positive,$0 < r < 8$.
Combining $r > 2$ and $r < 8$,we get $2 < r < 8$.

(A) The first circle lies in the first quadrant and touches both axes,so its centre $C_1$ is $(12, 12)$ and its radius $r_1 = 12$.
The second circle has centre $C_2 = (8, 9)$ and radius $r_2 = 7$.
The distance between the centres $C_1$ and $C_2$ is given by $d = \sqrt{(12 - 8)^2 + (12 - 9)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.
The difference between the radii is $|r_1 - r_2| = |12 - 7| = 5$.
Since the distance between the centres $d = |r_1 - r_2|$,the circles touch each other internally.

(D) Let the first circle be $C_1$ with center $O_1(3, 4)$ and radius $r_1 = r$.
Let the second circle be $C_2$ with center $O_2(0, 0)$ and radius $r_2 = R$.
For the circle $C_1$ to lie entirely within the circle $C_2$,the distance between their centers must be less than the difference of their radii.
That is,$d < R - r$,where $d$ is the distance between the centers $(3, 4)$ and $(0, 0)$.
$d = \sqrt{(3 - 0)^2 + (4 - 0)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
Thus,the condition is $5 < R - r$,or $R - r > 5$.

(A) The given circle is $x^2 + y^2 - 2x + 4y - 93 = 0$.
Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -1, f = 2, c = -93$.
The centre is $(-g, -f) = (1, -2)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{1 + 4 + 93} = \sqrt{98} = 7\sqrt{2}$.
The diameter of the circle is $2r = 14\sqrt{2}$.
Since the square is inscribed in the circle,its diagonal is equal to the diameter of the circle.
Let $l$ be the side length of the square. Then $l\sqrt{2} = 14\sqrt{2}$,which gives $l = 14$.
Since the sides are parallel to the coordinate axes,the distance of each side from the centre $(1, -2)$ is $l/2 = 7$.
The $x$-coordinates of the sides are $1 \pm 7$,i.e.,$x = -6$ and $x = 8$.
The $y$-coordinates of the sides are $-2 \pm 7$,i.e.,$y = -9$ and $y = 5$.
Thus,the vertices of the square are $(-6, -9), (-6, 5), (8, -9), (8, 5)$.

(A) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.
If the point $(m, \frac{1}{m})$ lies on this circle,then $m^2 + \frac{1}{m^2} + 2gm + \frac{2f}{m} + c = 0$.
Multiplying by $m^2$,we get the equation $m^4 + 2gm^3 + cm^2 + 2fm + 1 = 0$.
This is a fourth-degree equation in $m$ having $m_1, m_2, m_3, m_4$ as its roots.
By Vieta's formulas,the product of the roots is given by the constant term divided by the leading coefficient.
Therefore,$m_1 \cdot m_2 \cdot m_3 \cdot m_4 = \frac{1}{1} = 1$.

(C) The equation of the circle is $x^2 + y^2 - 4x - 2y - 11 = 0$.
Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -2$,$f = -1$,and $c = -11$.
The center of the circle is $(-g, -f) = (2, 1)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{4 + 1 + 11} = \sqrt{16} = 4$.
The length of the tangent $L$ from point $(x_1, y_1) = (4, 5)$ is given by $L = \sqrt{x_1^2 + y_1^2 - 4x_1 - 2y_1 - 11}$.
$L = \sqrt{4^2 + 5^2 - 4(4) - 2(5) - 11} = \sqrt{16 + 25 - 16 - 10 - 11} = \sqrt{4} = 2$.
The quadrilateral formed by the two tangents,the two radii,and the center is composed of two congruent right-angled triangles.
The area of the quadrilateral is $2 \times (\frac{1}{2} \times \text{base} \times \text{height}) = 2 \times (\frac{1}{2} \times L \times r) = L \times r$.
Area $= 2 \times 4 = 8 \text{ sq. units}$.

(C) The equation of the line passing through $C(-\sqrt{8}, \sqrt{8})$ with an angle of inclination $\theta = 135^\circ$ is given by $y - y_1 = m(x - x_1)$.
Here,$m = \tan(135^\circ) = -1$.
So,$y - \sqrt{8} = -1(x + \sqrt{8})$,which simplifies to $y - \sqrt{8} = -x - \sqrt{8}$,or $x + y = 0$.
The circle is given by $x = 5\cos\theta, y = 5\sin\theta$,which corresponds to $x^2 + y^2 = 25$.
The distance from the center $(0, 0)$ to the line $x + y = 0$ is $d = \frac{|0 + 0|}{\sqrt{1^2 + 1^2}} = 0$.
Since the distance from the center to the line is $0$,the line passes through the center of the circle.
Therefore,the line $x + y = 0$ is a diameter of the circle.
The length of the diameter $AB$ is $2r = 2 \times 5 = 10$.

(B) Given circles are $S_1 \equiv x^2 + y^2 = 2^2$ and $S_2 \equiv x^2 + y^2 - 6x - 8y - 24 = 0$.
For $S_2$,the centre is $(3, 4)$ and radius $r_2 = \sqrt{3^2 + 4^2 - (-24)} = \sqrt{9 + 16 + 24} = \sqrt{49} = 7$.
For $S_1$,the centre is $C_1 = (0, 0)$ and radius $r_1 = 2$.
The distance between the centres $C_1(0, 0)$ and $C_2(3, 4)$ is $d = \sqrt{(3-0)^2 + (4-0)^2} = \sqrt{9 + 16} = 5$.
Since $d = |r_2 - r_1| = |7 - 2| = 5$,the two circles touch each other internally.
When two circles touch internally,there is exactly $1$ common tangent.

(D) The equation of the circle is ${x^2} + {y^2} - px - qy = 0$.
Let the midpoint of a chord be $(h, 0)$ on the $x$-axis.
The equation of the chord with midpoint $(x_1, y_1)$ is $T = S_1$,where $T = xx_1 + yy_1 - \frac{p}{2}(x + x_1) - \frac{q}{2}(y + y_1)$ and $S_1 = x_1^2 + y_1^2 - px_1 - qy_1$.
Substituting $(x_1, y_1) = (h, 0)$,we get:
$xh - \frac{p}{2}(x + h) - \frac{q}{2}y = h^2 - ph$.
Since the chord passes through $(p, q)$,we substitute $x = p$ and $y = q$:
$ph - \frac{p}{2}(p + h) - \frac{q^2}{2} = h^2 - ph$.
Multiplying by $2$:
$2ph - p^2 - ph - q^2 = 2h^2 - 2ph$.
Rearranging terms:
$2h^2 - 3ph + p^2 + q^2 = 0$.
For two distinct chords to exist,the quadratic equation in $h$ must have two distinct real roots.
Thus,the discriminant $D > 0$:
$D = (-3p)^2 - 4(2)(p^2 + q^2) > 0$.
$9p^2 - 8p^2 - 8q^2 > 0$.
$p^2 - 8q^2 > 0$.
Therefore,$p^2 > 8q^2$.

(B) The first circle $C_1$ touches both axes and has a radius $r_1 = 2$. Thus,its center is $(2, 2)$.
Let the radius of the second circle $C_2$ be $a$. Since $C_2$ also touches both axes,its center is $(a, a)$.
Since $C_2$ touches $C_1$ externally,the distance between their centers must be equal to the sum of their radii:
$\sqrt{(a - 2)^2 + (a - 2)^2} = a + 2$
$\sqrt{2(a - 2)^2} = a + 2$
$\sqrt{2}|a - 2| = a + 2$
Since $a > 2$,we have $\sqrt{2}(a - 2) = a + 2$.
$a\sqrt{2} - 2\sqrt{2} = a + 2$
$a(\sqrt{2} - 1) = 2 + 2\sqrt{2}$
$a = \frac{2(\sqrt{2} + 1)}{\sqrt{2} - 1}$
Rationalizing the denominator:
$a = \frac{2(\sqrt{2} + 1)(\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = 2(2 + 1 + 2\sqrt{2}) = 2(3 + 2\sqrt{2}) = 6 + 4\sqrt{2}$.

(C) The tangent to the parabola $y = x^2$ at $(2, 4)$ is given by $\frac{1}{2}(y + 4) = x(2)$,which simplifies to $4x - y - 4 = 0$.
Since the circle touches the parabola at $(2, 4)$,the centre $(h, k)$ of the circle must lie on the normal to the parabola at $(2, 4)$.
The slope of the tangent is $4$,so the slope of the normal is $-\frac{1}{4}$.
The equation of the normal at $(2, 4)$ is $y - 4 = -\frac{1}{4}(x - 2)$,which simplifies to $x + 4y = 18$.
Thus,$h + 4k = 18$ $(i)$.
Since the circle passes through $(0, 1)$ and $(2, 4)$,the distance from the centre $(h, k)$ to these points must be equal:
$(h - 2)^2 + (k - 4)^2 = (h - 0)^2 + (k - 1)^2$.
Expanding this,$h^2 - 4h + 4 + k^2 - 8k + 16 = h^2 + k^2 - 2k + 1$,which simplifies to $4h + 6k = 19$ $(ii)$.
Solving equations $(i)$ and $(ii)$ simultaneously: from $(i)$,$h = 18 - 4k$. Substituting into $(ii)$ gives $4(18 - 4k) + 6k = 19$,so $72 - 16k + 6k = 19$,which means $10k = 53$,so $k = \frac{53}{10}$.
Then $h = 18 - 4(\frac{53}{10}) = 18 - \frac{106}{5} = \frac{90 - 106}{5} = -\frac{16}{5}$.
The centre is $\left( -\frac{16}{5}, \frac{53}{10} \right)$.

(D) To find the intersection,substitute $x^2 = ab - y^2$ into the ellipse equation: $\frac{ab - y^2}{a^2} + \frac{y^2}{b^2} = 1$.
This simplifies to $y^2(\frac{b^2 - a^2}{a^2b^2}) = 1 - \frac{b}{a} = \frac{a - b}{a}$.
Solving for $y^2$ and $x^2$,we get $y^2 = \frac{ab^2}{a + b}$ and $x^2 = \frac{a^2b}{a + b}$.
The point of intersection is $(x, y) = (a\sqrt{\frac{b}{a+b}}, b\sqrt{\frac{a}{a+b}})$.
The slope of the tangent to the ellipse is $m_1 = -\frac{b^2x}{a^2y} = -\frac{b^2}{a^2} \cdot \frac{a}{b} \sqrt{\frac{b}{a}} = -\frac{b}{a} \sqrt{\frac{b}{a}} = -\sqrt{\frac{b^3}{a^3}}$. Wait,simplifying: $m_1 = -\frac{b^2}{a^2} \cdot \frac{a\sqrt{b/(a+b)}}{b\sqrt{a/(a+b)}} = -\frac{b}{a} \sqrt{\frac{b}{a}}$.
The slope of the tangent to the circle is $m_2 = -\frac{x}{y} = -\frac{a\sqrt{b/(a+b)}}{b\sqrt{a/(a+b)}} = -\frac{a}{b} \sqrt{\frac{b}{a}} = -\sqrt{\frac{a}{b}}$.
The angle $\theta$ is given by $\tan \theta = |\frac{m_1 - m_2}{1 + m_1m_2}|$.
Substituting the values: $\tan \theta = |\frac{-\frac{b}{a}\sqrt{\frac{b}{a}} + \sqrt{\frac{a}{b}}}{1 + (\frac{b}{a}\sqrt{\frac{b}{a}})(\sqrt{\frac{a}{b}})}| = |\frac{-\frac{b}{a}\sqrt{\frac{b}{a}} + \sqrt{\frac{a}{b}}}{1 + \frac{b}{a}}| = |\frac{\frac{-b^2 + a^2}{a\sqrt{ab}}}{\frac{a+b}{a}}| = \frac{(a-b)(a+b)}{(a+b)\sqrt{ab}} = \frac{a-b}{\sqrt{ab}}$.
Thus,$\theta = \tan^{-1}\left(\frac{a - b}{\sqrt{ab}}\right)$.

(A) The given equation of the circle is $x^2 + y^2 - 3x + 8y - 4 = 0$.
Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $2g = -3 \Rightarrow g = -\frac{3}{2}$ and $2f = 8 \Rightarrow f = 4$.
The center of the circle is $(-g, -f) = (\frac{3}{2}, -4)$.
Let the other end of the diameter be $(h, k)$.
Since the center is the midpoint of the diameter,we have:
$\frac{6 + h}{2} = \frac{3}{2}$ $\Rightarrow 6 + h = 3$ $\Rightarrow h = -3$.
$\frac{-3 + k}{2} = -4$ $\Rightarrow -3 + k = -8$ $\Rightarrow k = -5$.
Thus,the other end of the diameter is $(-3, -5)$.

(C) For the first circle $x^2 + y^2 - 4x - 6y - 3 = 0$,the center $C_1 = (2, 3)$ and radius $r_1 = \sqrt{2^2 + 3^2 - (-3)} = \sqrt{4 + 9 + 3} = 4$.
For the second circle $x^2 + y^2 + 2x + 2y + 1 = 0$,the center $C_2 = (-1, -1)$ and radius $r_2 = \sqrt{(-1)^2 + (-1)^2 - 1} = \sqrt{1 + 1 - 1} = 1$.
The distance between the centers $C_1C_2 = \sqrt{(2 - (-1))^2 + (3 - (-1))^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5$.
Since $C_1C_2 = r_1 + r_2$ $(5 = 4 + 1)$,the two circles touch each other externally.
Therefore,the number of common tangents is $3$.

(A) For the first circle $x^2 + y^2 - 2x - 1 = 0$,the center $C_1 = (1, 0)$ and radius $r_1 = \sqrt{1^2 + 0^2 - (-1)} = \sqrt{2}$.
For the second circle $x^2 + y^2 - 2y - 7 = 0$,the center $C_2 = (0, 1)$ and radius $r_2 = \sqrt{0^2 + 1^2 - (-7)} = \sqrt{8} = 2\sqrt{2}$.
The distance between the centers $C_1$ and $C_2$ is $d = \sqrt{(1-0)^2 + (0-1)^2} = \sqrt{1 + 1} = \sqrt{2}$.
We observe that $|r_2 - r_1| = |2\sqrt{2} - \sqrt{2}| = \sqrt{2}$.
Since $d = |r_2 - r_1|$,the two circles touch each other internally.
When two circles touch internally,there is exactly $1$ common tangent.

(B) The perpendicular distance $d$ from the center $(0,0)$ to the line $mx - y + c = 0$ is given by:
$d = \frac{|m(0) - (0) + c|}{\sqrt{m^2 + (-1)^2}} = \frac{|c|}{\sqrt{1 + m^2}}$
In the right-angled triangle formed by the radius $a$,the half-chord $b$,and the perpendicular distance $d$,we have by Pythagoras theorem:
$a^2 = b^2 + d^2$
$d^2 = a^2 - b^2$
Substituting the value of $d$:
$\left(\frac{|c|}{\sqrt{1 + m^2}}\right)^2 = a^2 - b^2$
$\frac{c^2}{1 + m^2} = a^2 - b^2$
$c^2 = (1 + m^2)(a^2 - b^2)$

(C) The circle is $x^2 + y^2 = 5^2$,with center $O(0, 0)$ and radius $r = 5$.
Let $Q = (3, 4)$ and $R = (-4, 3)$.
The slope of $OQ$ is $m_1 = \frac{4-0}{3-0} = \frac{4}{3}$.
The slope of $OR$ is $m_2 = \frac{3-0}{-4-0} = -\frac{3}{4}$.
Since $m_1 \times m_2 = \frac{4}{3} \times (-\frac{3}{4}) = -1$,the lines $OQ$ and $OR$ are perpendicular.
Thus,the central angle $\angle QOR = \frac{\pi}{2}$.
By the circle theorem,the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.
Therefore,$\angle QPR = \frac{1}{2} \angle QOR = \frac{1}{2} \times \frac{\pi}{2} = \frac{\pi}{4}$.

(C) The given circle is $x^2 + y^2 - 4x + 2y - 11 = 0$.
Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -2$ and $f = 1$.
The center of the circle is $(-g, -f) = (2, -1)$.
The system of parallel chords is given by $x - 2y + c = 0$.
The diameter corresponding to this system of parallel chords is the line perpendicular to the chords and passing through the center of the circle.
The slope of the chords is $m = 1/2$.
The slope of the diameter is $m' = -1/m = -2$.
The equation of the diameter passing through $(2, -1)$ with slope $-2$ is:
$y - (-1) = -2(x - 2)$
$y + 1 = -2x + 4$
$2x + y - 3 = 0$.

(A) Given circle is $x^2 + y^2 - 2x = 0$ $(i)$ and the line is $y = x$ $(ii)$.
Substituting $y = x$ in $(i)$,we get $x^2 + x^2 - 2x = 0$,which simplifies to $2x^2 - 2x = 0$.
This gives $2x(x - 1) = 0$,so $x = 0$ or $x = 1$.
For $x = 0$,$y = 0$,so $A = (0, 0)$.
For $x = 1$,$y = 1$,so $B = (1, 1)$.
The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.
Substituting the points $(0, 0)$ and $(1, 1)$,we get $(x - 0)(x - 1) + (y - 0)(y - 1) = 0$.
This simplifies to $x(x - 1) + y(y - 1) = 0$,which is $x^2 - x + y^2 - y = 0$ or $x^2 + y^2 - x - y = 0$.

(B) The equation of the chord of a circle with a given midpoint $(x_1, y_1)$ is given by $T = S_1$.
Here,the circle is $S: x^2 + y^2 - 8x = 0$.
The midpoint is $(x_1, y_1) = (4, 3)$.
$T = x(4) + y(3) - 4(x + 4) = 4x + 3y - 4x - 16 = 3y - 16$.
$S_1 = (4)^2 + (3)^2 - 8(4) = 16 + 9 - 32 = -7$.
Equating $T = S_1$,we get $3y - 16 = -7$.
$3y = 9$,which simplifies to $y = 3$.

(C) The length of the tangent from a point $(x_1, y_1)$ to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is given by $\sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c}$.
Given the point $(3, -4)$ and the circle $x^2 + y^2 - 4x - 6y + 3 = 0$.
The square of the length of the tangent is $S_1 = x_1^2 + y_1^2 - 4x_1 - 6y_1 + 3$.
Substituting the values: $S_1 = (3)^2 + (-4)^2 - 4(3) - 6(-4) + 3$.
$S_1 = 9 + 16 - 12 + 24 + 3$.
$S_1 = 40$.

(D) The given circle equation is $x^2 + y^2 + 4y - 4 = 0$.
Comparing this with the general equation $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = 0$,$f = 2$,and $c = -4$.
The center of the circle is $(-g, -f) = (0, -2)$.
The radius $r$ is $\sqrt{g^2 + f^2 - c} = \sqrt{0^2 + 2^2 - (-4)} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$.
The perpendicular distance $p$ from the center $(0, -2)$ to the line $x - y + 1 = 0$ is calculated using the formula $p = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$.
$p = \frac{|1(0) - 1(-2) + 1|}{\sqrt{1^2 + (-1)^2}} = \frac{|0 + 2 + 1|}{\sqrt{2}} = \frac{3}{\sqrt{2}}$.
The length of the chord is given by $2\sqrt{r^2 - p^2}$.
Length $= 2\sqrt{(2\sqrt{2})^2 - (\frac{3}{\sqrt{2}})^2} = 2\sqrt{8 - \frac{9}{2}} = 2\sqrt{\frac{16 - 9}{2}} = 2\sqrt{\frac{7}{2}} = \sqrt{4 \cdot \frac{7}{2}} = \sqrt{14}$.

(D) For circle $C_1: x^2 + y^2 = 4$,center $C_1 = (0, 0)$ and radius $r_1 = 2$.
For circle $C_2: x^2 + y^2 - 6x - 8y - 24 = 0$,center $C_2 = (3, 4)$ and radius $r_2 = \sqrt{3^2 + 4^2 - (-24)} = \sqrt{9 + 16 + 24} = \sqrt{49} = 7$.
The distance between centers $|C_1C_2| = \sqrt{(3-0)^2 + (4-0)^2} = \sqrt{9 + 16} = 5$.
We check the condition $|C_1C_2|$ vs $r_1 + r_2$ and $|r_1 - r_2|$.
$r_1 + r_2 = 2 + 7 = 9$.
$|r_1 - r_2| = |2 - 7| = 5$.
Since $|C_1C_2| = |r_1 - r_2| = 5$,the circles touch each other internally.
When circles touch internally,they have only $1$ common tangent.
Thus,Statement $(A)$ is false.
Reason $(R)$ is a standard geometric theorem for circles,which is true.
Therefore,$A$ is false but $R$ is true.

(A) The given circle is $x^2 + y^2 = 4$,which has center $C = (0, 0)$ and radius $r = 2$.
The distance from the point $P(6, 8)$ to the center $C(0, 0)$ is given by the distance formula:
$d = \sqrt{(6-0)^2 + (8-0)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$.
The minimum distance from the point to the circle is $d - r = 10 - 2 = 8$.
The maximum distance from the point to the circle is $d + r = 10 + 2 = 12$.
Thus,the minimum and maximum distances are $8$ and $12$ respectively.

(C) The length of the tangent from a point $(x_1, y_1)$ to a circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is given by $\sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c}$.
For the first circle $x^2 + y^2 + x + y - 4 = 0$,the length $T_1 = \sqrt{1^2 + 2^2 + 1 + 2 - 4} = \sqrt{1 + 4 + 1 + 2 - 4} = \sqrt{4} = 2$.
For the second circle,we first normalize the equation: $x^2 + y^2 - \frac{1}{3}x - \frac{1}{3}y + \frac{k}{3} = 0$.
The length $T_2 = \sqrt{1^2 + 2^2 - \frac{1}{3}(1) - \frac{1}{3}(2) + \frac{k}{3}} = \sqrt{1 + 4 - \frac{1}{3} - \frac{2}{3} + \frac{k}{3}} = \sqrt{5 - 1 + \frac{k}{3}} = \sqrt{4 + \frac{k}{3}}$.
Given the ratio $\frac{T_1}{T_2} = \frac{4}{3}$,we have $\frac{2}{\sqrt{4 + k/3}} = \frac{4}{3}$.
Squaring both sides: $\frac{4}{4 + k/3} = \frac{16}{9} \Rightarrow 36 = 64 + \frac{16k}{3}$.
$36 - 64 = \frac{16k}{3}$ $\Rightarrow -28 = \frac{16k}{3}$ $\Rightarrow k = \frac{-28 \times 3}{16} = \frac{-7 \times 3}{4} = -\frac{21}{4}$.

(C) The lines are $y - \sqrt{3}x = 6$ and $y + \sqrt{3}x = 6$,and the $x$-axis is $y = 0$.
For $y = 0$,the lines intersect the $x$-axis at $x = -2\sqrt{3}$ and $x = 2\sqrt{3}$.
The intersection of the two lines $y - \sqrt{3}x = 6$ and $y + \sqrt{3}x = 6$ is $(0, 6)$.
The vertices of the triangle are $A(0, 6)$,$B(-2\sqrt{3}, 0)$,and $C(2\sqrt{3}, 0)$.
This is an isosceles triangle with base $BC$ on the $x$-axis.
The circumcenter $(0, k)$ must be equidistant from $(0, 6)$ and $(2\sqrt{3}, 0)$.
So,$|6 - k| = \sqrt{(2\sqrt{3} - 0)^2 + (0 - k)^2}$.
$(6 - k)^2 = 12 + k^2$.
$36 - 12k + k^2 = 12 + k^2$.
$12k = 24$,so $k = 2$.
The circumcenter is $(0, 2)$ and the radius $R = |6 - 2| = 4$.
The equation of the circle is $(x - 0)^2 + (y - 2)^2 = 4^2$.
$x^2 + y^2 - 4y + 4 = 16$.
$x^2 + y^2 - 4y = 12$.

(B) The equation of the circle is $x^2 + y^2 - 4x - 2y - 20 = 0$. The center $C$ is $(2, 1)$ and the radius $r$ is $\sqrt{2^2 + 1^2 - (-20)} = \sqrt{4 + 1 + 20} = \sqrt{25} = 5$.
The distance from point $P(10, 7)$ to the center $C(2, 1)$ is $PC = \sqrt{(10 - 2)^2 + (7 - 1)^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10$.
Since $PC > r$ $(10 > 5)$,the point $P$ lies outside the circle.
The maximum distance from point $P$ to the circle is given by $PC + r = 10 + 5 = 15$.

(A) Let the center of the circle be $(h, k)$. Since the circle touches the $X$-axis at $(3, 0)$,the $x$-coordinate of the center is $h = 3$ and the radius $r = |k|$.
Thus,the equation of the circle is $(x - 3)^2 + (y - k)^2 = k^2$.
This simplifies to $x^2 - 6x + 9 + y^2 - 2ky + k^2 = k^2$,or $x^2 + y^2 - 6x - 2ky + 9 = 0$.
The circle cuts an intercept of $8$ on the positive $Y$-axis. Setting $x = 0$,we get $y^2 - 2ky + 9 = 0$.
The length of the intercept on the $Y$-axis is given by $2\sqrt{f^2 - c}$,where $f = -k$ and $c = 9$.
So,$2\sqrt{(-k)^2 - 9} = 8 \implies \sqrt{k^2 - 9} = 4$.
Squaring both sides,$k^2 - 9 = 16 \implies k^2 = 25 \implies k = 5$ (since it is on the positive $Y$-axis).
Substituting $k = 5$ into the equation,we get $x^2 + y^2 - 6x - 2(5)y + 9 = 0$,which is $x^2 + y^2 - 6x - 10y + 9 = 0$.

(B) The radius $r$ of the circle is the distance between the center $(1, 2)$ and the point $(4, 6)$ on the circle.
Using the distance formula,$r = \sqrt{(4 - 1)^2 + (6 - 2)^2}$.
$r = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
The area of the circle is given by the formula $A = \pi r^2$.
Substituting $r = 5$,we get $A = \pi(5)^2 = 25\pi$.

(B) For the first circle $(x - 1)^2 + (y - 3)^2 = r^2$,the center $C_1 = (1, 3)$ and radius $R_1 = r$.
For the second circle $x^2 + y^2 - 8x + 2y + 8 = 0$,the center $C_2 = (-(-8)/2, -(2)/2) = (4, -1)$ and radius $R_2 = \sqrt{g^2 + f^2 - c} = \sqrt{(-4)^2 + (1)^2 - 8} = \sqrt{16 + 1 - 8} = \sqrt{9} = 3$.
The distance between the centers $d = \sqrt{(4 - 1)^2 + (-1 - 3)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = 5$.
Two circles intersect at two distinct points if $|R_1 - R_2| < d < R_1 + R_2$.
Substituting the values: $|r - 3| < 5 < r + 3$.
From $r + 3 > 5$,we get $r > 2$.
From $|r - 3| < 5$,we get $-5 < r - 3 < 5$,which implies $-2 < r < 8$. Since $r$ must be positive,$0 < r < 8$.
Combining $r > 2$ and $r < 8$,we get $2 < r < 8$.

(C) The line is given by $y - 1 = m(x - 1)$,which simplifies to $mx - y + (1 - m) = 0$.
For a line $Ax + By + C = 0$ to intersect the circle $x^2 + y^2 = a^2$ at two distinct points,the perpendicular distance from the center $(0, 0)$ to the line must be less than the radius $a = 2$.
The distance $d = \frac{|m(0) - (0) + (1 - m)|}{\sqrt{m^2 + (-1)^2}} = \frac{|1 - m|}{\sqrt{m^2 + 1}}$.
Setting $d < 2$,we get $\frac{|1 - m|}{\sqrt{m^2 + 1}} < 2$.
Squaring both sides: $(1 - m)^2 < 4(m^2 + 1)$.
$1 - 2m + m^2 < 4m^2 + 4$.
$3m^2 + 2m + 3 > 0$.
Since the discriminant of $3m^2 + 2m + 3$ is $D = 2^2 - 4(3)(3) = 4 - 36 = -32 < 0$,the quadratic $3m^2 + 2m + 3$ is always positive for all real values of $m$.
Thus,there are infinitely many values of $m$ for which the line intersects the circle at two distinct points.

(B) The given equation of the circle is $x^2 + y^2 - 4x - 4y + 4 = 0$.
Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -2$,$f = -2$,and $c = 4$.
The center of the circle is $(-g, -f) = (2, 2)$.
The radius of the circle is $r = \sqrt{g^2 + f^2 - c} = \sqrt{(-2)^2 + (-2)^2 - 4} = \sqrt{4 + 4 - 4} = \sqrt{4} = 2$.
Since the distance from the center $(2, 2)$ to both the $x$-axis and $y$-axis is equal to the radius $r = 2$,the circle touches both axes.

(C) The equation of the circle is $2x^2 + 2y^2 = 3$.
Dividing by $2$,we get $x^2 + y^2 = \frac{3}{2}$,or $x^2 + y^2 - \frac{3}{2} = 0$.
The length of the tangent from a point $(x_1, y_1)$ to a circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is given by $\sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c}$.
Substituting $(1, 5)$ into the equation $x^2 + y^2 - \frac{3}{2} = 0$:
Length $= \sqrt{1^2 + 5^2 - \frac{3}{2}} = \sqrt{1 + 25 - 1.5} = \sqrt{26 - 1.5} = \sqrt{24.5}$.
$\sqrt{24.5} = \sqrt{\frac{49}{2}} = \frac{7}{\sqrt{2}} = \frac{7\sqrt{2}}{2}$.

(D) The given equation of the circle is $2x^2 + 2y^2 + 3x + 4y - 1 = 0$.
Dividing by $2$,we get $x^2 + y^2 + \frac{3}{2}x + 2y - \frac{1}{2} = 0$.
The center of the circle $(h, k')$ is given by $(-\frac{g}{2}, -\frac{f}{2})$,where $2g = \frac{3}{2}$ and $2f = 2$.
Thus,the center is $(-\frac{3}{4}, -1)$.
Since the diameter $y = 2x + k$ passes through the center of the circle,we substitute the center coordinates into the equation:
$-1 = 2(-\frac{3}{4}) + k$
$-1 = -\frac{3}{2} + k$
$k = -1 + \frac{3}{2} = \frac{1}{2}$.

(C) The radius $r$ of the incircle of an equilateral triangle with side $a = 6$ is given by $r = \frac{a}{2\sqrt{3}} = \frac{6}{2\sqrt{3}} = \sqrt{3}$.
The square is inscribed in this circle,so the diagonal of the square is equal to the diameter of the circle.
Diagonal of the square $d = 2r = 2\sqrt{3}$.
Let the side of the square be $x$. Then $x\sqrt{2} = d = 2\sqrt{3}$.
$x = \frac{2\sqrt{3}}{\sqrt{2}} = \sqrt{2} \times \sqrt{3} = \sqrt{6}$.
Area of the square $= x^2 = (\sqrt{6})^2 = 6$.

(B) The given circle is $x^2 + y^2 = 8$,which can be written as $x^2 + y^2 = (2\sqrt{2})^2$.
Here,the radius $r = 2\sqrt{2}$.
The director circle of a circle $x^2 + y^2 = r^2$ is $x^2 + y^2 = 2r^2$.
Substituting $r^2 = 8$,we get the equation of the director circle as $x^2 + y^2 = 2(8) = 16$.

(C) Let the center of the circle be $(h, k)$ and radius be $r$.
Since the circle touches the $Y$-axis at $(0, 4)$,the distance from the center to the $Y$-axis is the radius,so $|h| = r$.
Given the circle touches at $(0, 4)$,the $y$-coordinate of the center must be $k = 4$ or $k = -4$.
Thus,the center is $(\pm r, \pm 4)$.
The equation of the circle is $(x \mp r)^{2} + (y \mp 4)^{2} = r^{2}$.
Since it cuts an intercept of $6$ units on the $X$-axis,the length of the intercept is $2\sqrt{r^{2} - k^{2}} = 6$.
$\sqrt{r^{2} - 4^{2}} = 3 \implies r^{2} - 16 = 9 \implies r^{2} = 25 \implies r = 5$.
Substituting $r=5$ and $k=\pm 4$,the center is $(\pm 5, \pm 4)$.
The equation of the circle is $(x \pm 5)^{2} + (y \pm 4)^{2} = 25$.