Tangent and normal to a circle Questions in English

(C) The equation of the line is $3x + y + k = 0$.
The equation of the circle is $x^{2} + y^{2} = 10$,which has its center at $(0, 0)$ and radius $r = \sqrt{10}$.
If a line $Ax + By + C = 0$ is tangent to a circle with center $(x_{1}, y_{1})$ and radius $r$,then the perpendicular distance from the center to the line must equal the radius:
$\left| \frac{Ax_{1} + By_{1} + C}{\sqrt{A^{2} + B^{2}}} \right| = r$
Substituting the values:
$\left| \frac{3(0) + 1(0) + k}{\sqrt{3^{2} + 1^{2}}} \right| = \sqrt{10}$
$\left| \frac{k}{\sqrt{9 + 1}} \right| = \sqrt{10}$
$\left| \frac{k}{\sqrt{10}} \right| = \sqrt{10}$
$|k| = \sqrt{10} \times \sqrt{10}$
$|k| = 10$
Therefore,$k = \pm 10$.

(D) The equation of the circle is $x^2+y^2+2gx+2fy+c=0$.
The center of the circle is $C(-g, -f)$.
The normal to a circle at any point $P(x_1, y_1)$ always passes through the center $C(-g, -f)$.
The slope of the normal is the slope of the line segment $CP$.
The slope $m$ is given by $\frac{y_2-y_1}{x_2-x_1} = \frac{y_1 - (-f)}{x_1 - (-g)} = \frac{y_1+f}{x_1+g}$.
Thus,the slope of the normal is $\frac{y_1+f}{x_1+g}$.

(B) Given $h^2 + hk + k^2 = 37$ ... $(i)$
Radius $r = \sqrt{10}$.
The equation of the circle $S$ is $(x - h)^2 + (y - k)^2 = 10$ ... $(ii)$
Since $(1, 2)$ lies on the circle,$(1 - h)^2 + (2 - k)^2 = 10$.
Expanding this,$1 - 2h + h^2 + 4 - 4k + k^2 = 10$,which simplifies to $h^2 + k^2 - 2h - 4k = 5$.
Substituting $h^2 + k^2 = 37 - hk$ from $(i)$ into this equation:
$37 - hk - 2h - 4k = 5 \Rightarrow hk + 2h + 4k = 32$ ... $(iii)$
The perpendicular distance from the centre $(h, k)$ to the tangent line $3x + y - 5 = 0$ is equal to the radius $\sqrt{10}$:
$\frac{|3h + k - 5|}{\sqrt{3^2 + 1^2}} = \sqrt{10} \Rightarrow |3h + k - 5| = 10$.
Assuming $3h + k - 5 = 10$,we get $3h + k = 15 \Rightarrow k = 15 - 3h$.
Substituting $k = 15 - 3h$ into $(iii)$:
$h(15 - 3h) + 2h + 4(15 - 3h) = 32$
$15h - 3h^2 + 2h + 60 - 12h = 32$
$-3h^2 + 5h + 28 = 0 \Rightarrow 3h^2 - 5h - 28 = 0$.
Solving the quadratic equation $(3h + 7)(h - 4) = 0$,we get $h = 4$ or $h = -7/3$.
If $h = 4$,then $k = 15 - 3(4) = 3$.
Thus,$k = 3$.

(A) The given circle is $x^2+y^2-4x-8y+16=0$. Its center is $C(2, 4)$ and radius $r = \sqrt{2^2+4^2-16} = 2$.
Let $P(2+\sqrt{3}, 3)$ be the point of tangency.
The slope of the radius $CP$ is $m_{CP} = \frac{3-4}{(2+\sqrt{3})-2} = \frac{-1}{\sqrt{3}}$.
The tangent is perpendicular to the radius,so the slope of the tangent is $m_t = \sqrt{3}$.
Since the circle rolls along the tangent by $2$ units,the new center $C'$ is at a distance of $2$ units from $C$ in the direction of the tangent.
The angle of the tangent with the $x$-axis is $\theta = \tan^{-1}(\sqrt{3}) = 60^{\circ}$.
The new center $C'$ is $(2+2\cos 60^{\circ}, 4+2\sin 60^{\circ}) = (2+2(1/2), 4+2(\sqrt{3}/2)) = (3, 4+\sqrt{3})$.
The radius remains $r=2$. The equation of the new circle is $(x-3)^2 + (y-(4+\sqrt{3}))^2 = 2^2$.
Expanding this: $x^2-6x+9 + y^2-2(4+\sqrt{3})y + (4+\sqrt{3})^2 = 4$.
$x^2+y^2-6x-2(4+\sqrt{3})y + 9 + 16 + 3 + 8\sqrt{3} - 4 = 0$.
$x^2+y^2-6x-2(4+\sqrt{3})y + (24+8\sqrt{3}) = 0$.

(A) The normals to a circle always intersect at its center. Given the normals $(x-1)(y-2)=0$,the center of the circle is $(1, 2)$.
Since $3x+4y=6$ is a tangent to the circle,the radius $r$ is the perpendicular distance from the center $(1, 2)$ to the line $3x+4y-6=0$.
$r = \frac{|3(1) + 4(2) - 6|}{\sqrt{3^2 + 4^2}} = \frac{|3 + 8 - 6|}{\sqrt{9 + 16}} = \frac{5}{5} = 1$.
The equation of the circle with center $(h, k) = (1, 2)$ and radius $r = 1$ is $(x-h)^2 + (y-k)^2 = r^2$.
$(x-1)^2 + (y-2)^2 = 1^2$,which simplifies to $(x-1)^2 + (y-2)^2 = 1$.
Thus,option $A$ is correct.

(C) The equation of the circle is $x^2 + y^2 - 6x + 8y + k = 0$. The centre $C$ is $(3, -4)$ and the radius $r$ is $\sqrt{3^2 + (-4)^2 - k} = \sqrt{25 - k}$.
Since $3x + 4y - 43 = 0$ is a tangent at $P$,the distance from $C(3, -4)$ to the line is equal to the radius $r$.
$r = \frac{|3(3) + 4(-4) - 43|}{\sqrt{3^2 + 4^2}} = \frac{|9 - 16 - 43|}{5} = \frac{|-50|}{5} = 10$.
Thus,$r^2 = 100$,so $25 - k = 100$,which gives $k = -75$.
The circle is $S \equiv x^2 + y^2 - 6x + 8y - 75 = 0$.
Point $Q$ divides $CP$ in the ratio $-1:2$. Let $C = (3, -4)$ and $P = (x_p, y_p)$.
$Q = \frac{-1(P) + 2(C)}{-1 + 2} = 2C - P = 2(3, -4) - P = (6 - x_p, -8 - y_p)$.
The power of point $Q$ with respect to the circle $S=0$ is $S(Q) = x_Q^2 + y_Q^2 - 6x_Q + 8y_Q - 75$.
Since $P$ lies on the circle,$x_p^2 + y_p^2 - 6x_p + 8y_p - 75 = 0$.
Substituting $Q$ into the circle equation: $(6-x_p)^2 + (-8-y_p)^2 - 6(6-x_p) + 8(-8-y_p) - 75 = 0$.
$36 - 12x_p + x_p^2 + 64 + 16y_p + y_p^2 - 36 + 6x_p - 64 - 8y_p - 75 = (x_p^2 + y_p^2 - 6x_p + 8y_p - 75) = 0$ is not correct here.
Actually,the power of point $Q$ is $CQ^2 - r^2$. Since $Q$ divides $CP$ in ratio $-1:2$,$CQ = |-1| \times CP / |2-1| = r$.
Thus $CQ^2 = r^2$,so the power of point $Q$ is $r^2 - r^2 = 0$.

(C) The equation of the circle is $x^2 + y^2 - 6x + 4y - 12 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -3, f = 2, c = -12$. The center is $(-g, -f) = (3, -2)$.
Since the line $4x - 3y + 7 = 0$ is tangent to the circle at $(\alpha, \beta)$,the radius passing through the point of contact is perpendicular to the tangent.
The slope of the tangent is $m_1 = \frac{4}{3}$.
The slope of the normal (radius) passing through $(3, -2)$ and $(\alpha, \beta)$ is $m_2 = \frac{\beta - (-2)}{\alpha - 3} = \frac{\beta + 2}{\alpha - 3}$.
Since $m_1 \times m_2 = -1$,we have $\frac{4}{3} \times \frac{\beta + 2}{\alpha - 3} = -1$,which implies $4(\beta + 2) = -3(\alpha - 3)$ $\Rightarrow 4\beta + 8 = -3\alpha + 9$ $\Rightarrow 3\alpha + 4\beta = 1$.
Also,the point $(\alpha, \beta)$ lies on the line $4\alpha - 3\beta + 7 = 0$.
Solving the system:
$3\alpha + 4\beta = 1$ $(i)$
$4\alpha - 3\beta = -7$ (ii)
Multiplying $(i)$ by $3$ and (ii) by $4$: $9\alpha + 12\beta = 3$ and $16\alpha - 12\beta = -28$.
Adding these,$25\alpha = -25 \Rightarrow \alpha = -1$.
Substituting $\alpha = -1$ into $(i)$: $3(-1) + 4\beta = 1$ $\Rightarrow 4\beta = 4$ $\Rightarrow \beta = 1$.
Thus,$\alpha + 2\beta = -1 + 2(1) = 1$.

(D) The given equation of the circle is $x^2+y^2+2x-12y-132=0$.
Completing the square,we get $(x+1)^2+(y-6)^2 = 132+1+36 = 169 = 13^2$.
Thus,the center is $(-1, 6)$ and the radius $r = 13$.
The slope of the line $12x+5y+k=0$ is $m_1 = -\frac{12}{5}$.
The slope of the tangent perpendicular to this line is $m = \frac{5}{12}$.
The equation of the tangent can be written as $5x-12y+c = 0$.
The perpendicular distance from the center $(-1, 6)$ to the tangent is equal to the radius $13$:
$\frac{|5(-1)-12(6)+c|}{\sqrt{5^2+(-12)^2}} = 13$
$\frac{|-5-72+c|}{13} = 13$
$|c-77| = 169$
$c-77 = 169 \Rightarrow c = 246$ or $c-77 = -169 \Rightarrow c = -92$.
Thus,the equations of the tangents are $5x-12y+246=0$ and $5x-12y-92=0$.

(B) The given equation of the circle is $x^2 + y^2 + 2gx + 2fy + d = 0$.
Its centre is $(-g, -f)$.
$A$ normal to a circle always passes through its centre.
The given equation of the line is $ax + by + c = 0$.
Since the line is a normal,it must pass through the centre $(-g, -f)$.
Substituting the centre $(-g, -f)$ into the line equation:
$a(-g) + b(-f) + c = 0$
$-ag - bf + c = 0$
Multiplying by $-1$,we get:
$ag + bf - c = 0$

(C) Let the slope of the tangent be $m$. The equation of the line passing through $(3, 4)$ with slope $m$ is $y - 4 = m(x - 3)$,which simplifies to $mx - y + (4 - 3m) = 0$.
Since this line is a tangent to the circle $x^2 + y^2 = 9$ (with center $(0, 0)$ and radius $r = 3$),the perpendicular distance from the center to the line must equal the radius.
Using the formula $d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$,we have $3 = \frac{|m(0) - 1(0) + 4 - 3m|}{\sqrt{m^2 + (-1)^2}}$.
$3\sqrt{m^2 + 1} = |4 - 3m|$.
Squaring both sides: $9(m^2 + 1) = (4 - 3m)^2$.
$9m^2 + 9 = 16 - 24m + 9m^2$.
$9 = 16 - 24m$.
$24m = 7$.
$m = \frac{7}{24}$.

(C) Let the angle between the tangents be $2\theta = \cos^{-1}\left(\frac{7}{16}\right)$.
Then $\cos(2\theta) = \frac{7}{16}$.
Using the identity $\cos(2\theta) = 2\cos^2\theta - 1$,we get $2\cos^2\theta - 1 = \frac{7}{16}$,which implies $2\cos^2\theta = \frac{23}{16}$,so $\cos^2\theta = \frac{23}{32}$.
Thus,$\sin^2\theta = 1 - \frac{23}{32} = \frac{9}{32}$,and $\tan^2\theta = \frac{9/32}{23/32} = \frac{9}{23}$.
For a circle $x^2 + y^2 + 4x + 4y + c = 0$,the center is $O(-2, -2)$ and radius $r = \sqrt{(-2)^2 + (-2)^2 - c} = \sqrt{8 - c}$.
The distance from point $P(2, 2)$ to center $O(-2, -2)$ is $d = \sqrt{(2 - (-2))^2 + (2 - (-2))^2} = \sqrt{4^2 + 4^2} = \sqrt{32}$.
In the right-angled triangle formed by the point,center,and point of tangency,$\tan\theta = \frac{r}{\sqrt{d^2 - r^2}}$.
So,$\tan^2\theta = \frac{r^2}{d^2 - r^2} = \frac{8 - c}{32 - (8 - c)} = \frac{8 - c}{24 + c}$.
Equating the two expressions for $\tan^2\theta$: $\frac{8 - c}{24 + c} = \frac{9}{23}$.
$23(8 - c) = 9(24 + c) \implies 184 - 23c = 216 + 9c$.
$-32c = 32 \implies c = -1$.
Since the problem states two such circles exist,there might be a misinterpretation of the geometry or parameters,but based on the provided equation,the sum of values of $c$ is $-1$.

(B) The equation of the circle is $x^2+y^2-4x+6y-4=0$.
Comparing this with the general form $x^2+y^2+2gx+2fy+c=0$,we get $g=-2$ and $f=3$.
The center of the circle is $(-g, -f) = (2, -3)$.
The normal to a circle at any point $(x_1, y_1)$ always passes through the center of the circle.
Thus,the normal is the line passing through $(1, 1)$ and $(2, -3)$.
The slope of this line is $m = \frac{-3-1}{2-1} = \frac{-4}{1} = -4$.
The equation of the line is $(y-1) = -4(x-1)$.
$y-1 = -4x+4$.
$4x+y = 5$.

(C) The equation of the circle is $x^2 + y^2 = 1$.
The equation of the tangent to the circle $x^2 + y^2 = r^2$ at the point $(x_1, y_1)$ is given by $x x_1 + y y_1 = r^2$.
Here,$r^2 = 1$,so the tangent at $(x_1, y_1)$ is $x x_1 + y y_1 = 1$.
We are given that the line $\frac{x}{a} + \frac{y}{b} = 1$ is a tangent to the circle.
Comparing this with $x x_1 + y y_1 = 1$,we get $x_1 = \frac{1}{a}$ and $y_1 = \frac{1}{b}$.
Since $(x_1, y_1)$ is the point of tangency,it must lie on the circle $x^2 + y^2 = 1$.
Therefore,the point $\left(\frac{1}{a}, \frac{1}{b}\right)$ lies on the circle.

(C) The equation of the circle is $S: x^2+y^2+6x+6y-2=0$.
Point $Q$ lies on the $Y$-axis and on the line $5x-2y+6=0$.
Setting $x=0$ in the line equation: $5(0)-2y+6=0$ $\Rightarrow -2y=-6$ $\Rightarrow y=3$.
So,the coordinates of $Q$ are $(0, 3)$.
The length of the tangent from a point $(x_1, y_1)$ to a circle $S=0$ is given by $\sqrt{S_1}$.
Substituting $Q(0, 3)$ into the circle equation $S(x, y) = x^2+y^2+6x+6y-2$:
$S_1 = 0^2 + 3^2 + 6(0) + 6(3) - 2 = 0 + 9 + 0 + 18 - 2 = 25$.
Therefore,the length of the tangent $PQ = \sqrt{S_1} = \sqrt{25} = 5$.

(A) The distance $PC$ between the center $C(1, 2)$ and the point $P(16, 7)$ is given by:
$PC = \sqrt{(16-1)^2 + (7-2)^2} = \sqrt{15^2 + 5^2} = \sqrt{225 + 25} = \sqrt{250}$.
The quadrilateral $PACB$ consists of two congruent right-angled triangles $\triangle PAC$ and $\triangle PBC$. The area of $PACB$ is $2 \times \text{Area}(\triangle PAC) = 2 \times (\frac{1}{2} \times AP \times r) = AP \times r = 75$.
Let $AP = x$,so $x = \frac{75}{r}$.
In right-angled triangle $\triangle PAC$,by the Pythagorean theorem:
$x^2 + r^2 = PC^2$
$(\frac{75}{r})^2 + r^2 = 250$
$\frac{5625}{r^2} + r^2 = 250$
$r^4 - 250r^2 + 5625 = 0$.
Let $u = r^2$,then $u^2 - 250u + 5625 = 0$.
$(u - 225)(u - 25) = 0$.
Thus,$r^2 = 225$ or $r^2 = 25$.
If $r^2 = 225$,then $r = 15$. Then $x = \frac{75}{15} = 5$. In this case,$PC^2 = 15^2 + 5^2 = 250$,which is consistent.
If $r^2 = 25$,then $r = 5$. Then $x = \frac{75}{5} = 15$. In this case,$PC^2 = 5^2 + 15^2 = 250$,which is also consistent.
Since $5$ is an option,the radius is $5$.

(B) Since $(\alpha, \beta)$ lies on the line $3x - 4y = 1$,we have $3\alpha - 4\beta = 1$,which implies $\beta = \frac{3\alpha - 1}{4}$.
Since $(\alpha, \beta)$ also lies on the circle $(x - 1)^2 + (y + 2)^2 = 4$,we substitute $\beta$:
$(\alpha - 1)^2 + (\frac{3\alpha - 1}{4} + 2)^2 = 4$
$(\alpha - 1)^2 + (\frac{3\alpha + 7}{4})^2 = 4$
$16(\alpha - 1)^2 + (3\alpha + 7)^2 = 64$
$16(\alpha^2 - 2\alpha + 1) + (9\alpha^2 + 42\alpha + 49) = 64$
$16\alpha^2 - 32\alpha + 16 + 9\alpha^2 + 42\alpha + 49 = 64$
$25\alpha^2 + 10\alpha + 1 = 0$
$(5\alpha + 1)^2 = 0$
$\alpha = -\frac{1}{5}$.
Substituting $\alpha$ into the line equation: $\beta = \frac{3(-\frac{1}{5}) - 1}{4} = \frac{-\frac{3}{5} - 1}{4} = \frac{-\frac{8}{5}}{4} = -\frac{2}{5}$.
Thus,$(\alpha, \beta) = (-\frac{1}{5}, -\frac{2}{5})$.

(A) The equation of a line perpendicular to $y=mx+1$ is of the form $x+my+k=0$,or $y=-\frac{1}{m}x+c$.
For a circle $x^2+y^2=r^2$,the condition for the line $y=m_1x+c$ to be a tangent is $c^2=r^2(1+m_1^2)$.
Here,the slope of the tangent is $m_1 = -\frac{1}{m}$ and $r=1$.
Substituting these into the condition: $c^2 = 1 \cdot (1 + (-\frac{1}{m})^2) = 1 + \frac{1}{m^2} = \frac{m^2+1}{m^2}$.
Thus,$c = \pm \frac{\sqrt{m^2+1}}{m}$.
The equation of the tangent is $y = -\frac{1}{m}x \pm \frac{\sqrt{m^2+1}}{m}$.
Multiplying by $m$,we get $my = -x \pm \sqrt{m^2+1}$,which simplifies to $x+my \pm \sqrt{1+m^2}=0$.

(B) The slope of the tangent $m = \tan 60^{\circ} = \sqrt{3}$.
The equation of the tangent to the circle $x^2 + y^2 = a^2$ with slope $m$ is given by $y = mx \pm a\sqrt{1+m^2}$.
Here,the radius $a = \sqrt{9} = 3$.
Substituting the values,we get $y = \sqrt{3}x \pm 3\sqrt{1+(\sqrt{3})^2}$.
$y = \sqrt{3}x \pm 3\sqrt{1+3}$.
$y = \sqrt{3}x \pm 3(2)$.
$y = \sqrt{3}x \pm 6$.
Rearranging the terms,we get $\sqrt{3}x - y \pm 6 = 0$.

(A) line passing through the point $(4,0)$ with slope $m$ is given by $y - 0 = m(x - 4)$,which simplifies to $mx - y - 4m = 0$.
For this line to be a tangent to the circle $x^2 + y^2 = 4$ (with center $(0,0)$ and radius $r = 2$),the perpendicular distance from the center to the line must equal the radius.
Using the distance formula $d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$,we have $\frac{|m(0) - (0) - 4m|}{\sqrt{m^2 + (-1)^2}} = 2$.
$| -4m | = 2\sqrt{m^2 + 1}$.
Squaring both sides: $16m^2 = 4(m^2 + 1)$ $\Rightarrow 4m^2 = m^2 + 1$ $\Rightarrow 3m^2 = 1$ $\Rightarrow m = \pm \frac{1}{\sqrt{3}}$.
Substituting $m$ back into the line equation: $y = \pm \frac{1}{\sqrt{3}}(x - 4)$.

(B) The given circle is $x^2+y^2+4x+8y-38=0$.
Completing the square,we get $(x+2)^2+(y+4)^2=58$.
The center of the circle is $C(-2,-4)$.
Let $Q(x_1, y_1)$ be the other end of the diameter passing through $P(-9,-1)$.
Since $C$ is the midpoint of $PQ$,we have $\frac{x_1-9}{2}=-2 \Rightarrow x_1=5$ and $\frac{y_1-1}{2}=-4 \Rightarrow y_1=-7$.
Thus,$Q$ is $(5,-7)$.
The tangent at $Q$ is parallel to the tangent at $P$.
The slope of the radius $CP$ is $m_{CP} = \frac{-1-(-4)}{-9-(-2)} = \frac{3}{-7} = -\frac{3}{7}$.
The slope of the tangent at $P$ (and $Q$) is the negative reciprocal of the slope of the radius,which is $m = -\frac{1}{-3/7} = \frac{7}{3}$.
The equation of the tangent at $Q(5,-7)$ is $y - (-7) = \frac{7}{3}(x - 5)$.
$3(y+7) = 7(x-5)$ $\Rightarrow 3y+21 = 7x-35$ $\Rightarrow 7x-3y=56$.

(B) Given,the equation of the circle is $5x^2 + 5y^2 = 1$.
Dividing by $5$,we get $x^2 + y^2 = \frac{1}{5} = \left(\frac{1}{\sqrt{5}}\right)^2$.
Thus,the center is $(0, 0)$ and the radius $r = \frac{1}{\sqrt{5}}$.
The given line is $3x + 4y = 1$,which has a slope $m = -\frac{3}{4}$.
The equation of a tangent to the circle $x^2 + y^2 = r^2$ with slope $m$ is given by $y = mx \pm r\sqrt{1 + m^2}$.
Substituting $m = -\frac{3}{4}$ and $r = \frac{1}{\sqrt{5}}$:
$y = -\frac{3}{4}x \pm \frac{1}{\sqrt{5}} \sqrt{1 + \left(-\frac{3}{4}\right)^2}$
$y = -\frac{3}{4}x \pm \frac{1}{\sqrt{5}} \sqrt{1 + \frac{9}{16}}$
$y = -\frac{3}{4}x \pm \frac{1}{\sqrt{5}} \sqrt{\frac{25}{16}}$
$y = -\frac{3}{4}x \pm \frac{1}{\sqrt{5}} \cdot \frac{5}{4}$
$y = -\frac{3}{4}x \pm \frac{\sqrt{5}}{4}$
Multiplying by $4$:
$4y = -3x \pm \sqrt{5}$
$3x + 4y = \pm \sqrt{5}$.

(A) The equation of the circle is $x^2 + y^2 - x - 3y - 4 = 0$.
To find the slope of the tangent at $(1, 1)$,we differentiate the equation with respect to $x$:
$2x + 2yy' - 1 - 3y' = 0$
$y'(2y - 3) = 1 - 2x$
$y' = \frac{1 - 2x}{2y - 3}$
At the point $(1, 1)$,the slope of the tangent $m_T$ is:
$m_T = \frac{1 - 2(1)}{2(1) - 3} = \frac{-1}{-1} = 1$.
The slope of the normal $m_N$ is given by $m_N = -\frac{1}{m_T} = -\frac{1}{1} = -1$.
The equation of the normal at $(1, 1)$ is:
$y - 1 = -1(x - 1)$
$y - 1 = -x + 1$
$x + y - 2 = 0$.

(D) The equation of the circle is $x^2 + y^2 = 10$,so its center is $(0, 0)$ and radius $r = \sqrt{10}$.
Since the line $3x + y + k = 0$ is a tangent to the circle,the perpendicular distance from the center $(0, 0)$ to the line must be equal to the radius $r$.
The perpendicular distance from $(0, 0)$ to $3x + y + k = 0$ is given by $\left| \frac{3(0) + 1(0) + k}{\sqrt{3^2 + 1^2}} \right| = \left| \frac{k}{\sqrt{10}} \right|$.
Equating this to the radius: $\left| \frac{k}{\sqrt{10}} \right| = \sqrt{10}$.
$\Rightarrow |k| = \sqrt{10} \times \sqrt{10} = 10$.
Therefore,$k = \pm 10$.

(C) Let the equation of the line passing through $(4, -2)$ be $y - (-2) = m(x - 4)$,which simplifies to $mx - y - (4m + 2) = 0$.
For this line to be a tangent to the circle $x^2 + y^2 = 10$,the perpendicular distance from the center $(0, 0)$ to the line must be equal to the radius $r = \sqrt{10}$.
Using the distance formula $d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$,we have $\sqrt{10} = \frac{|m(0) - 1(0) - (4m + 2)|}{\sqrt{m^2 + (-1)^2}}$.
$\sqrt{10} = \frac{|4m + 2|}{\sqrt{m^2 + 1}}$.
Squaring both sides: $10(m^2 + 1) = (4m + 2)^2$.
$10m^2 + 10 = 16m^2 + 16m + 4$.
$6m^2 + 16m - 6 = 0$,which simplifies to $3m^2 + 8m - 3 = 0$.
$(3m - 1)(m + 3) = 0$,so $m = \frac{1}{3}$ or $m = -3$.
For $m = \frac{1}{3}$,the line is $y + 2 = \frac{1}{3}(x - 4) \implies 3y + 6 = x - 4 \implies x - 3y = 10$.
For $m = -3$,the line is $y + 2 = -3(x - 4) \implies y + 2 = -3x + 12 \implies 3x + y = 10$.
Thus,the equations are $3x + y = 10$ and $x - 3y = 10$.

(C) The equation of the circle is $x^2+y^2=16$,which has its center at $C(0,0)$.
Any normal to a circle always passes through its center.
Therefore,the normal at the point $P\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$ is the line passing through $C(0,0)$ and $P\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$.
The slope $m$ of the line $CP$ is given by $m = \frac{\frac{1}{\sqrt{3}} - 0}{\frac{1}{\sqrt{3}} - 0} = 1$.
The equation of the line passing through $(0,0)$ with slope $m=1$ is $y - 0 = 1(x - 0)$,which simplifies to $x - y = 0$.
Thus,option $C$ is correct.

(B) The distance between two parallel lines $y=mx+k_1$ and $y=mx+k_2$ is given by $d = \frac{|k_1-k_2|}{\sqrt{1+m^2}}$.
Since these lines are parallel tangents to a circle of radius $r=2$,the distance between them must be equal to the diameter of the circle,which is $2r = 2 \times 2 = 4$.
Here,$m = \sqrt{3}$,so $m^2 = 3$.
Substituting these values into the distance formula:
$\frac{|k_1-k_2|}{\sqrt{1+3}} = 4$
$\frac{|k_1-k_2|}{\sqrt{4}} = 4$
$\frac{|k_1-k_2|}{2} = 4$
$|k_1-k_2| = 8$.
Thus,option $B$ is correct.

(D) The mid-point $P$ of the line joining the origin $(0, 0)$ and the point $(4, -4)$ is given by $P = (\frac{0+4}{2}, \frac{0-4}{2}) = (2, -2)$.
The equation of the circle is $2x^2 + 2y^2 - y = 0$. Dividing by $2$,we get $x^2 + y^2 - \frac{1}{2}y = 0$.
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 $x_1 = 2, y_1 = -2, g = 0, f = -\frac{1}{4}, c = 0$:
Length $= \sqrt{(2)^2 + (-2)^2 - \frac{1}{2}(-2)} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \text{ units.}$

(C) Let the point be $P = (6, 8)$ and the equation of the circle be $S: x^2 + y^2 - 4 = 0$.
The length of the tangent from a point $(x_1, y_1)$ to a circle $S = 0$ is given by $\sqrt{S_1} = \sqrt{x_1^2 + y_1^2 - 4}$.
Substituting the coordinates $(6, 8)$ into the expression:
$\text{Length} = \sqrt{6^2 + 8^2 - 4} = \sqrt{36 + 64 - 4} = \sqrt{100 - 4} = \sqrt{96}$.
Simplifying $\sqrt{96} = \sqrt{16 \times 6} = 4 \sqrt{6}$.
Hence,option $C$ is correct.

(A) The equation of the tangent to the circle $x^2+y^2=r^2$ at the point $(x_1, y_1)$ is given by $xx_1+yy_1=r^2$.
Given the circle $x^2+y^2=25$,we have $r^2=25$.
The point of tangency is $(x_1, y_1) = (-3, 4)$.
Substituting these values into the formula:
$x(-3) + y(4) = 25$
$-3x + 4y = 25$
Rearranging the terms,we get:
$3x - 4y + 25 = 0$.

(B) The given equation of the circle is $x^2+y^2-2x+4y=0$.
Comparing this with the general form $x^2+y^2+2gx+2fy+c=0$,we get $g=-1$ and $f=2$.
The center of the circle is $C(-g, -f) = (1, -2)$.
Let $A(3, -1)$ be one end of the diameter and $B(x_1, y_1)$ be the other end.
Since the center $C$ is the midpoint of the diameter $AB$,we have $(1, -2) = (\frac{3+x_1}{2}, \frac{-1+y_1}{2})$.
Solving for $B$,we get $3+x_1=2 \implies x_1=-1$ and $-1+y_1=-4 \implies y_1=-3$.
So,the other end of the diameter is $B(-1, -3)$.
The tangent at $B(-1, -3)$ is perpendicular to the radius $CB$.
The slope of the radius $CB$ is $m_{CB} = \frac{-3-(-2)}{-1-1} = \frac{-1}{-2} = \frac{1}{2}$.
The slope of the tangent at $B$ is $m = -\frac{1}{m_{CB}} = -2$.
The equation of the tangent at $B(-1, -3)$ is $y - (-3) = -2(x - (-1))$.
$y+3 = -2x-2$,which simplifies to $2x+y+5=0$.

(C) The equation of the line at $(1,1)$ is $x+y-2=0$. The slope of this line is $-1$.
Since the normal is perpendicular to the tangent,the slope of the normal is $1$.
Thus,$\tan \theta = 1$,which gives $\theta = \frac{\pi}{4}$.
Let the centre of the circle be $(h, k)$. The coordinates of the centre are given by $h = 1 \pm r \cos \theta$ and $k = 1 \pm r \sin \theta$.
Given $r = \sqrt{2}$ and $\theta = \frac{\pi}{4}$,we have:
$h = 1 \pm \sqrt{2} \cos \frac{\pi}{4} = 1 \pm \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 1 \pm 1$.
$k = 1 \pm \sqrt{2} \sin \frac{\pi}{4} = 1 \pm \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 1 \pm 1$.
So,the possible centres are $(2, 2)$ or $(0, 0)$.
The equations of the circles are $x^2 + y^2 = 2$ or $(x-2)^2 + (y-2)^2 = 2$.
For the circle $x^2 + y^2 - 2 = 0$,the length of the tangent from $(1, 2)$ is $\sqrt{1^2 + 2^2 - 2} = \sqrt{1+4-2} = \sqrt{3}$.
For the circle $(x-2)^2 + (y-2)^2 - 2 = 0$,the length of the tangent from $(1, 2)$ is $\sqrt{(1-2)^2 + (2-2)^2 - 2} = \sqrt{(-1)^2 + 0^2 - 2} = \sqrt{1-2} = \sqrt{-1}$,which is not possible.
Thus,the length of the tangent is $\sqrt{3}$.

(A) The given equation of the circle is $x^2+y^2-2x-2y-3=0$.
Comparing this with the general equation $x^2+y^2+2gx+2fy+c=0$,we get $g=-1$ and $f=-1$.
The centre of the circle is $O(-g, -f) = (1, 1)$.
Since the normal at any point on a circle always passes through the centre,the line segment $PQ$ is a diameter of the circle.
Therefore,the centre $O(1, 1)$ is the midpoint of the diameter $PQ$.
Let the coordinates of $Q$ be $(x, y)$.
Using the midpoint formula,we have:
$\frac{-1+x}{2} = 1 \implies -1+x = 2 \implies x = 3$
$\frac{2+y}{2} = 1 \implies 2+y = 2 \implies y = 0$
Thus,the coordinates of $Q$ are $(3, 0)$.

(B) The given circle is $x^2+y^2+6x+4y-3=0$.
Comparing this with the general equation $x^2+y^2+2gx+2fy+c=0$,we get $g=3$ and $f=2$.
The center of the circle is $(-g, -f) = (-3, -2)$.
The normal to a circle at any point always passes through the center of the circle.
The normal is a line passing through the center $(-3, -2)$ and the point $(1, -2)$.
Since the $y$-coordinates of both points are the same $(-2)$,the line is a horizontal line given by $y = -2$.
Thus,the equation of the normal is $y+2=0$.

(D) The equation of the circle is $x^2+y^2-6x+4y+12=0$.
Completing the square,we get $(x-3)^2+(y+2)^2 = 3^2+(-2)^2-12 = 9+4-12 = 1$.
So,the center $O$ is $(3,-2)$ and the radius $r$ is $1$.
The distance $d$ from the point $A(2,3)$ to the center $O(3,-2)$ is $d = \sqrt{(3-2)^2+(-2-3)^2} = \sqrt{1^2+(-5)^2} = \sqrt{26}$.
Let $\alpha$ be the half-angle between the tangents. In the right-angled triangle formed by the point $A$,the center $O$,and the point of tangency $P$,we have $\sin(\alpha) = \frac{r}{d} = \frac{1}{\sqrt{26}}$.
Then $\cos(\alpha) = \sqrt{1-\sin^2(\alpha)} = \sqrt{1-\frac{1}{26}} = \sqrt{\frac{25}{26}} = \frac{5}{\sqrt{26}}$.
Thus,$\tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)} = \frac{1/\sqrt{26}}{5/\sqrt{26}} = \frac{1}{5}$.
The total angle between the tangents is $\theta = 2\alpha$.
Using the formula $\tan(2\alpha) = \frac{2\tan(\alpha)}{1-\tan^2(\alpha)}$,we get $\tan(\theta) = \frac{2(1/5)}{1-(1/5)^2} = \frac{2/5}{24/25} = \frac{2}{5} \times \frac{25}{24} = \frac{5}{12}$.
Therefore,$\theta = \tan^{-1}\left(\frac{5}{12}\right)$.

(A) The locus of the point of intersection of perpendicular tangents to a circle is called the director circle.
For the circle $x^2+y^2=a^2$,the equation of the director circle is $x^2+y^2=2a^2$.
Since the point $(10,4)$ lies on the director circle,we have:
$10^2+4^2 = 2a^2$
$100+16 = 2a^2$
$116 = 2a^2$
$a^2 = 58$
$a = \sqrt{58}$

(C) The equation of the pair of tangents from a point $(x_1, y_1)$ to a circle $S = 0$ is given by $SS_1 = T^2$.
Given $S = x^2 + y^2 + 2x + 2y + 1 = 0$ and point $(1, 1)$.
$S_1 = 1^2 + 1^2 + 2(1) + 2(1) + 1 = 1 + 1 + 2 + 2 + 1 = 7$.
$T = x(1) + y(1) + (x + 1) + (y + 1) + 1 = 2x + 2y + 3$.
Substituting these into $SS_1 = T^2$:
$7(x^2 + y^2 + 2x + 2y + 1) = (2x + 2y + 3)^2$.
$7x^2 + 7y^2 + 14x + 14y + 7 = 4x^2 + 4y^2 + 9 + 8xy + 12x + 12y$.
Rearranging the terms:
$3x^2 - 8xy + 3y^2 + 2x + 2y - 2 = 0$.

(C) The equation of the circle is $S \equiv x^2+y^2-4x-8y+4=0$. The center is $(2, 4)$ and the radius $r = \sqrt{2^2+4^2-4} = \sqrt{16} = 4$.
Let $d$ be the distance from the origin $(0,0)$ to the center $(2,4)$,so $d = \sqrt{2^2+4^2} = \sqrt{20} = 2\sqrt{5}$.
Let the angle between the tangents be $\alpha$. Then the angle between the line joining the origin to the center and one of the tangents is $\alpha/2$.
In the right-angled triangle formed by the origin,the center,and the point of tangency,we have $\sin(\alpha/2) = r/d = 4/(2\sqrt{5}) = 2/\sqrt{5}$.
Thus,$\cos(\alpha/2) = \sqrt{1 - \sin^2(\alpha/2)} = \sqrt{1 - 4/5} = 1/\sqrt{5}$.
Therefore,$\tan(\alpha/2) = \sin(\alpha/2) / \cos(\alpha/2) = (2/\sqrt{5}) / (1/\sqrt{5}) = 2$.
Using the formula $\tan \alpha = \frac{2 \tan(\alpha/2)}{1 - \tan^2(\alpha/2)}$,we get $\tan \alpha = \frac{2(2)}{1 - 2^2} = \frac{4}{1-4} = -4/3$.
Since $\alpha$ is an acute angle,we take the magnitude,so $\tan \alpha = 4/3$.

(A) The equation of the circle is $(x+\lambda)^2+(y+1)^2=\lambda^2$.
Here,the center $C$ is $(-\lambda, -1)$ and the radius $r$ is $|\lambda|$.
Let $O$ be the origin $(0,0)$ and $P$ be the point of contact of a tangent from $O$ to the circle.
The angle between the tangents is $\frac{\pi}{2}$,so the angle $\angle COP = \frac{1}{2} \times \frac{\pi}{2} = \frac{\pi}{4} = 45^{\circ}$.
In the right-angled triangle $\triangle OCP$,we have $\tan(\angle COP) = \frac{CP}{OP} = \frac{r}{OP}$.
Since $\angle COP = 45^{\circ}$,$\tan(45^{\circ}) = 1$,so $OP = CP = |\lambda|$.
Also,the distance $OC = \sqrt{(-\lambda-0)^2 + (-1-0)^2} = \sqrt{\lambda^2+1}$.
By the Pythagorean theorem in $\triangle OCP$,$OC^2 = OP^2 + CP^2$.
Substituting the values,$(\sqrt{\lambda^2+1})^2 = |\lambda|^2 + |\lambda|^2$.
$\lambda^2 + 1 = 2\lambda^2$.
Therefore,$\lambda^2 = 1$.

(A) Given the circle equation $x^2+y^2=50$ and the line $x+7=0$.
Substituting $x = -7$ into the circle equation:
$(-7)^2 + y^2 = 50$
$49 + y^2 = 50$
$y^2 = 1 \Rightarrow y = \pm 1$.
Thus,the points of intersection are $P_1(-7, 1)$ and $P_2(-7, -1)$.
The equation of the tangent to the circle $x^2+y^2=r^2$ at $(x_1, y_1)$ is given by $xx_1 + yy_1 = r^2$.
For point $(-7, 1)$: $-7x + y = 50 \Rightarrow 7x - y + 50 = 0$.
For point $(-7, -1)$: $-7x - y = 50 \Rightarrow 7x + y + 50 = 0$.
Therefore,the required equations are $7x+y+50=0$ and $7x-y+50=0$.

(A) The given equation of the circle is $x^2+y^2+4x+4y-1=0$.
Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=2, f=2, c=-1$.
The center of the circle is $O(-g, -f) = (-2, -2)$ and the radius $r = \sqrt{g^2+f^2-c} = \sqrt{4+4-(-1)} = \sqrt{9} = 3$.
Let the external point be $C(1, 1)$. The distance $OC$ is $\sqrt{(1 - (-2))^2 + (1 - (-2))^2} = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}$.
Let $\alpha$ be the angle between the line joining the center to the external point and the tangent. In the right-angled triangle $\triangle OAC$ (where $A$ is the point of tangency),$\sin \alpha = \frac{OA}{OC} = \frac{r}{OC} = \frac{3}{3\sqrt{2}} = \frac{1}{\sqrt{2}}$.
Thus,$\alpha = 45^\circ$ or $\frac{\pi}{4}$.
The angle between the pair of tangents is $2\alpha = 2 \times \frac{\pi}{4} = \frac{\pi}{2}$.

(C) The given equation of the circle is $x^2+y^2-14x+2y+25=0$.
Completing the square,we get $(x^2-14x+49) + (y^2+2y+1) - 49 - 1 + 25 = 0$,which simplifies to $(x-7)^2 + (y+1)^2 = 25 = 5^2$.
Thus,the radius $r = 5$ and the center $P = (7, -1)$.
The distance from the origin $O(0,0)$ to the center $P(7,-1)$ is $OP = \sqrt{7^2 + (-1)^2} = \sqrt{49+1} = \sqrt{50} = 5\sqrt{2}$.
Let the tangents from the origin touch the circle at $A$ and $B$. In the right-angled triangle $\triangle OAP$,$\sin \theta = \frac{AP}{OP} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$.
Therefore,$\theta = 45^{\circ}$.
Similarly,for $\triangle OBP$,$\sin \alpha = \frac{BP}{OP} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$,so $\alpha = 45^{\circ}$.
The total angle between the tangents is $\theta + \alpha = 45^{\circ} + 45^{\circ} = 90^{\circ}$.

(A) The equation of the circle is $x^2+y^2-2x+4y-11=0$.
Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-1, f=2, c=-11$.
The radius $r = \sqrt{g^2+f^2-c} = \sqrt{(-1)^2+2^2-(-11)} = \sqrt{1+4+11} = \sqrt{16} = 4$.
The length of the tangent $L_T$ from point $(1,3)$ is $\sqrt{S_1} = \sqrt{1^2+3^2-2(1)+4(3)-11} = \sqrt{1+9-2+12-11} = \sqrt{9} = 3$.
Let the angle between the pair of tangents be $2\theta$.
In the right-angled triangle formed by the center,the point $(1,3)$,and the point of contact,$\tan\theta = \frac{r}{L_T} = \frac{4}{3}$.
We know that $\sin(2\theta) = \frac{2\tan\theta}{1+\tan^2\theta} = \frac{2(4/3)}{1+(4/3)^2} = \frac{8/3}{1+16/9} = \frac{8/3}{25/9} = \frac{8}{3} \times \frac{9}{25} = \frac{24}{25}$.
Therefore,the angle $2\theta = \sin^{-1}\left(\frac{24}{25}\right)$.

(A) The equation of the given circle is $x^2+y^2-2x-4y-20=0$ ... $(i)$.
The centre $A$ is $(1, 2)$ and the radius $r = \sqrt{1^2+2^2+20} = \sqrt{25} = 5$.
The equation of the tangent at $B(1, 7)$ is $x(1) + y(7) - (x+1) - 2(y+7) - 20 = 0$,which simplifies to $5y = 35$,or $y = 7$ ... $(ii)$.
The equation of the tangent at $D(4, -2)$ is $x(4) + y(-2) - (x+4) - 2(y-2) - 20 = 0$,which simplifies to $3x - 4y = 20$ ... $(iii)$.
Solving $(ii)$ and $(iii)$ for the intersection point $C$: Substituting $y=7$ into $(iii)$,$3x - 4(7) = 20$ $\Rightarrow 3x = 48$ $\Rightarrow x = 16$. So,$C = (16, 7)$.
The area of the quadrilateral $ABCD$ is $2 \times \text{Area}(\triangle ABC) = 2 \times (\frac{1}{2} \times r \times L)$,where $L$ is the length of the tangent from $C$ to the circle.
$L = \sqrt{S_1} = \sqrt{16^2 + 7^2 - 2(16) - 4(7) - 20} = \sqrt{256 + 49 - 32 - 28 - 20} = \sqrt{225} = 15$.
Area $= r \times L = 5 \times 15 = 75$ square units.
Thus,option $(A)$ is correct.

(D) The equation of the circle is $x^2+y^2-2x-4y=0$.
Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-1$,$f=-2$,and $c=0$.
The center of the circle is $C(-g, -f) = (1, 2)$ and the radius $r = \sqrt{g^2+f^2-c} = \sqrt{(-1)^2+(-2)^2-0} = \sqrt{1+4} = \sqrt{5}$.
Let $P$ be the point $(4,3)$. The distance $d$ from $P$ to the center $C(1,2)$ is $d = \sqrt{(4-1)^2+(3-2)^2} = \sqrt{3^2+1^2} = \sqrt{9+1} = \sqrt{10}$.
Let $\theta$ be the angle between the tangents. The angle between the radius and the tangent is $\alpha = \frac{\theta}{2}$.
In the right-angled triangle formed by the center,the point $P$,and the point of tangency,we have $\sin(\alpha) = \frac{r}{d} = \frac{\sqrt{5}}{\sqrt{10}} = \frac{1}{\sqrt{2}}$.
Thus,$\alpha = \frac{\pi}{4}$.
The angle between the tangents is $\theta = 2\alpha = 2 \times \frac{\pi}{4} = \frac{\pi}{2}$.

(A) Given circle: $x^2+y^2+4x-6y+4=0$.
Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=2, f=-3, c=4$.
Centre $O = (-g, -f) = (-2, 3)$.
Radius $r = \sqrt{g^2+f^2-c} = \sqrt{4+9-4} = \sqrt{9} = 3$.
Distance from origin $P(0,0)$ to centre $O(-2,3)$ is $d = \sqrt{(-2-0)^2 + (3-0)^2} = \sqrt{4+9} = \sqrt{13}$.
Let $\alpha$ be the angle between the tangent and the line joining the origin to the centre.
In the right-angled triangle formed by the origin,the point of tangency,and the centre,$\sin \alpha = \frac{r}{d} = \frac{3}{\sqrt{13}}$.
Then $\cos \alpha = \sqrt{1-\sin^2 \alpha} = \sqrt{1-\frac{9}{13}} = \sqrt{\frac{4}{13}} = \frac{2}{\sqrt{13}}$.
Thus,$\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{3/\sqrt{13}}{2/\sqrt{13}} = \frac{3}{2}$.
The angle between the two tangents is $2\alpha = 2 \tan^{-1} \left(\frac{3}{2}\right)$.

(C) For the circle $C_1: x^2+y^2-4x+12y-216=0$,the center $C_1$ is $(2, -6)$ and radius $r_1 = \sqrt{2^2 + (-6)^2 - (-216)} = \sqrt{4 + 36 + 216} = \sqrt{256} = 16$.
For the circle $C_2: x^2+y^2+6x-12y+36=0$,the center $C_2$ is $(-3, 6)$ and radius $r_2 = \sqrt{(-3)^2 + 6^2 - 36} = \sqrt{9 + 36 - 36} = 3$.
The distance between centers $d = \sqrt{(2 - (-3))^2 + (-6 - 6)^2} = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = 13$.
Since $r_1 - r_2 = 16 - 3 = 13$,which is equal to $d$,the circles touch each other internally.
There is only one common tangent at the point of contact.
The point of contact $P$ divides the line segment $C_1C_2$ externally in the ratio $r_1 : r_2 = 16 : 3$.
$P = \left( \frac{16(-3) - 3(2)}{16 - 3}, \frac{16(6) - 3(-6)}{16 - 3} \right) = \left( \frac{-48 - 6}{13}, \frac{96 + 18}{13} \right) = \left( -\frac{54}{13}, \frac{114}{13} \right)$.
The slope of the line $C_1C_2$ is $m_{C_1C_2} = \frac{6 - (-6)}{-3 - 2} = \frac{12}{-5} = -\frac{12}{5}$.
The common tangent is perpendicular to the line joining the centers.
Therefore,the slope of the common tangent is $m = -\frac{1}{m_{C_1C_2}} = -\frac{1}{-12/5} = \frac{5}{12}$.

(D) Let $S_1: x^2+y^2+2x-2y-2=0$. The center $C_1 = (-1, 1)$ and radius $r_1 = \sqrt{(-1)^2 + 1^2 - (-2)} = \sqrt{4} = 2$.
Let $S_2: x^2+y^2-2x+2y+1=0$. The center $C_2 = (1, -1)$ and radius $r_2 = \sqrt{1^2 + (-1)^2 - 1} = \sqrt{1} = 1$.
Since the circles are external to each other,the common tangents intersect at a point $P$ which divides the line segment joining the centers $C_1$ and $C_2$ externally in the ratio of their radii $r_1 : r_2 = 2 : 1$.
Using the section formula for external division,the coordinates of $P$ are:
$P = \left( \frac{2(1) - 1(-1)}{2-1}, \frac{2(-1) - 1(1)}{2-1} \right) = \left( \frac{3}{1}, \frac{-3}{1} \right) = (3, -3)$.
Let the equation of the tangent line passing through $P(3, -3)$ be $y + 3 = m(x - 3)$,which simplifies to $mx - y - 3m - 3 = 0$.
The perpendicular distance from the center $C_2(1, -1)$ to this line must equal the radius $r_2 = 1$.
$\left| \frac{m(1) - (-1) - 3m - 3}{\sqrt{m^2 + (-1)^2}} \right| = 1$ $\Rightarrow \left| \frac{-2m - 2}{\sqrt{m^2 + 1}} \right| = 1$.
Squaring both sides: $\frac{4(m+1)^2}{m^2+1} = 1 \Rightarrow 4(m^2 + 2m + 1) = m^2 + 1$.
$4m^2 + 8m + 4 = m^2 + 1 \Rightarrow 3m^2 + 8m + 3 = 0$.
The product of the slopes $m_1 m_2$ is given by the constant term divided by the coefficient of $m^2$ in the quadratic equation $3m^2 + 8m + 3 = 0$.
Therefore,$m_1 m_2 = \frac{3}{3} = 1$.

(A) The given equation of the line is $(x-2) \cos \theta + (y-2) \sin \theta = 1$.
This line represents a tangent to the circle for all values of $\theta$.
The distance from the center $(h, k)$ of the circle to the tangent line must be equal to the radius $r$.
Rewriting the line as $(x-2) \cos \theta + (y-2) \sin \theta - 1 = 0$,the distance from $(h, k)$ is $\frac{|(h-2) \cos \theta + (k-2) \sin \theta - 1|}{\sqrt{\cos^2 \theta + \sin^2 \theta}} = r$.
Since this holds for all $\theta$,we must have $h-2 = 0$ and $k-2 = 0$,which gives the center $(2, 2)$.
Then,the distance becomes $|-1| = r$,so $r = 1$.
The equation of the circle is $(x-2)^2 + (y-2)^2 = 1^2$.
Expanding this,we get $x^2 - 4x + 4 + y^2 - 4y + 4 = 1$.
Thus,$x^2 + y^2 - 4x - 4y + 7 = 0$.