Tangent and normal to a circle Questions in English

(B) The given line is $yy' = 2ax + 2ax'$,which can be rewritten as $2ax - yy' + 2ax' = 0$.
Comparing this with the standard form $Ax + By + C = 0$,the slope $m_1 = \frac{2a}{y'}$.
Since the required line is perpendicular to this line,its slope $m_2$ must satisfy $m_1 \times m_2 = -1$.
Thus,$m_2 = -\frac{y'}{2a}$.
The equation of the line passing through $(x', y')$ with slope $m_2$ is $y - y' = -\frac{y'}{2a}(x - x')$.
Multiplying by $2a$,we get $2ay - 2ay' = -y'(x - x')$.
$2ay - 2ay' = -xy' + x'y'$.
Rearranging the terms,we get $xy' + 2ay - 2ay' - x'y' = 0$.

(A) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$ $(i)$.
Since it passes through $(-1, -3)$,we have $1 - 3 + 2g(-1) + 2f(-3) + c = 0 \Rightarrow 10 - 2g - 6f + c = 0$ $(ii)$.
Since it passes through $(3, 0)$,we have $9 + 0 + 2g(3) + 0 + c = 0 \Rightarrow 9 + 6g + c = 0$ $(iii)$.
The centre of the circle is $C(-g, -f)$. The line $4x + 3y - 12 = 0$ is tangent at $(3, 0)$,so the normal at $(3, 0)$ passes through the centre $(-g, -f)$.
The slope of the tangent is $-4/3$,so the slope of the normal is $3/4$.
Thus,$\frac{-f - 0}{-g - 3} = \frac{3}{4}$ $\Rightarrow -4f = -3g - 9$ $\Rightarrow 3g - 4f + 9 = 0$ $(iv)$.
Solving equations $(ii)$,$(iii)$,and $(iv)$:
From $(iii)$,$c = -9 - 6g$. Substituting into $(ii)$: $10 - 2g - 6f - 9 - 6g = 0$ $\Rightarrow 1 - 8g - 6f = 0$ $\Rightarrow 8g + 6f = 1$.
We have the system: $3g - 4f = -9$ and $8g + 6f = 1$.
Multiplying the first by $3$ and the second by $2$: $9g - 12f = -27$ and $16g + 12f = 2$.
Adding them: $25g = -25 \Rightarrow g = -1$.
Substituting $g = -1$ into $3g - 4f = -9$: $-3 - 4f = -9$ $\Rightarrow -4f = -6$ $\Rightarrow f = 3/2$.
Substituting $g = -1$ into $c = -9 - 6g$: $c = -9 - 6(-1) = -3$.
The equation is $x^2 + y^2 - 2x + 3y - 3 = 0$.

(B) 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 $\sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c}$.
Given the point $(5, 3)$ and the circle $x^2 + y^2 + 2x + ky + 17 = 0$,the length of the tangent is $7$.
Substituting the values into the formula:
$\sqrt{5^2 + 3^2 + 2(5) + k(3) + 17} = 7$
$\sqrt{25 + 9 + 10 + 3k + 17} = 7$
$\sqrt{61 + 3k} = 7$
Squaring both sides:
$61 + 3k = 49$
$3k = 49 - 61$
$3k = -12$
$k = -4$.

(B) The condition for the line $y = Mx + C$ to be a tangent to the circle $x^2 + y^2 = a^2$ is $C^2 = a^2(1 + M^2)$.
Given the line $lx + my + n = 0$,we can rewrite it as $y = -\frac{l}{m}x - \frac{n}{m}$.
Here,the slope $M = -\frac{l}{m}$ and the intercept $C = -\frac{n}{m}$.
Substituting these into the condition $C^2 = a^2(1 + M^2)$,we get:
$(-\frac{n}{m})^2 = a^2(1 + (-\frac{l}{m})^2)$
$\frac{n^2}{m^2} = a^2(1 + \frac{l^2}{m^2})$
$\frac{n^2}{m^2} = a^2(\frac{m^2 + l^2}{m^2})$
$n^2 = a^2(l^2 + m^2)$.

(D) Let the line passing through the origin $(0, 0)$ be $y = mx$,or $mx - y = 0$.
This line is a tangent to the circle $(x - 7)^2 + (y + 1)^2 = 25$ if the perpendicular distance from the center $(7, -1)$ to the line is equal to the radius $r = 5$.
Using the formula for the distance from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$,we have $\frac{|m(7) - (-1)|}{\sqrt{m^2 + (-1)^2}} = 5$.
$\frac{|7m + 1|}{\sqrt{m^2 + 1}} = 5$.
Squaring both sides: $(7m + 1)^2 = 25(m^2 + 1)$.
$49m^2 + 14m + 1 = 25m^2 + 25$.
$24m^2 + 14m - 24 = 0$.
$12m^2 + 7m - 12 = 0$.
This is a quadratic equation in $m$. Let the roots be $m_1$ and $m_2$. The product of the slopes is $m_1 m_2 = \frac{-12}{12} = -1$.
Since the product of the slopes is $-1$,the two tangents are perpendicular to each other.
Therefore,the angle between the two tangents is $\frac{\pi}{2}$.

(A) The condition for the line $y = mx + c$ to be a tangent to the circle $x^2 + y^2 = a^2$ is $c^2 = a^2(1 + m^2)$.
Substituting $y = mx + c$ into the circle equation: $x^2 + (mx + c)^2 = a^2$.
$x^2 + m^2x^2 + 2mxc + c^2 = a^2$.
$(1 + m^2)x^2 + 2mxc + (c^2 - a^2) = 0$.
Since the line is a tangent,the discriminant must be zero,leading to the x-coordinate of the point of contact: $x = -\frac{2mc}{2(1 + m^2)} = -\frac{mc}{1 + m^2}$.
Using $c^2 = a^2(1 + m^2)$,we have $1 + m^2 = \frac{c^2}{a^2}$.
Thus,$x = -\frac{mc}{c^2/a^2} = -\frac{a^2m}{c}$.
Substituting $x$ back into $y = mx + c$: $y = m(-\frac{a^2m}{c}) + c = \frac{-a^2m^2 + c^2}{c} = \frac{a^2(1 + m^2) - a^2m^2}{c} = \frac{a^2}{c}$.
Therefore,the point of contact is $\left( -\frac{a^2m}{c}, \frac{a^2}{c} \right)$.

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

(D) The equation of the circle is $x^2 + y^2 = 16$,which has center $(0, 0)$ and radius $r = \sqrt{16} = 4$.
For the line $y = mx + c$ to touch the circle $x^2 + y^2 = r^2$,the condition is $c^2 = r^2(1 + m^2)$.
Here,$m = \sqrt{3}$,$c = k$,and $r = 4$.
Substituting these values,we get $k^2 = 4^2(1 + (\sqrt{3})^2)$.
$k^2 = 16(1 + 3) = 16 \times 4 = 64$.
Therefore,$k = \pm 8$.

(C) The equation of the circle is $x^2 + y^2 = 36$,so the radius $a = 6$.
Since the tangent is inclined at an angle of $45^\circ$ to the $x$-axis,the slope $m = \tan(45^\circ) = 1$.
The equation of a tangent to the circle $x^2 + y^2 = a^2$ with slope $m$ is given by $y = mx \pm a\sqrt{1 + m^2}$.
Substituting $m = 1$ and $a = 6$:
$y = 1(x) \pm 6\sqrt{1 + (1)^2}$
$y = x \pm 6\sqrt{2}$.

(C) The equation of the normal to the circle $x^2 + y^2 = a^2$ at any point $(x_1, y_1)$ on the circle passes through the center $(0, 0)$.
Since the normal passes through the center $(0, 0)$ and the point $P = \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)$,its slope $m$ is given by $m = \frac{y_1 - 0}{x_1 - 0} = \frac{1/\sqrt{2}}{1/\sqrt{2}} = 1$.
The equation of the line passing through $(0, 0)$ with slope $m = 1$ is $y - 0 = 1(x - 0)$,which simplifies to $y = x$ or $x - y = 0$.

(A) The equation of the circle is $x^2 + y^2 - 22x - 4y + 25 = 0$.
Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -11, f = -2, c = 25$.
The center is $(-g, -f) = (11, 2)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{121 + 4 - 25} = \sqrt{100} = 10$.
$A$ line perpendicular to $5x + 12y + 8 = 0$ is of the form $12x - 5y + k = 0$.
Since this line is a tangent to the circle,the perpendicular distance from the center $(11, 2)$ to the line must equal the radius $r = 10$.
$\frac{|12(11) - 5(2) + k|}{\sqrt{12^2 + (-5)^2}} = 10$
$\frac{|132 - 10 + k|}{13} = 10$
$|122 + k| = 130$
$122 + k = 130$ or $122 + k = -130$
$k = 8$ or $k = -252$.
Thus,the equations of the tangents are $12x - 5y + 8 = 0$ and $12x - 5y - 252 = 0$.

(D) The line $x \cos \alpha + y \sin \alpha - p = 0$ is a tangent to the circle if the perpendicular distance from the center to the line is equal to the radius of the circle.
The given circle equation is $x^2 + y^2 - 2ax \cos \alpha - 2ay \sin \alpha = 0$.
Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get center $(g, f) = (a \cos \alpha, a \sin \alpha)$ and radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{a^2 \cos^2 \alpha + a^2 \sin^2 \alpha - 0} = a$.
The perpendicular distance from $(a \cos \alpha, a \sin \alpha)$ to the line $x \cos \alpha + y \sin \alpha - p = 0$ is given by:
$d = \frac{|(a \cos \alpha) \cos \alpha + (a \sin \alpha) \sin \alpha - p|}{\sqrt{\cos^2 \alpha + \sin^2 \alpha}}$
$d = |a(\cos^2 \alpha + \sin^2 \alpha) - p| = |a - p|$.
Setting $d = r$,we get $|a - p| = a$.
This implies $a - p = a$ or $a - p = -a$.
Therefore,$p = 0$ or $p = 2a$.

(C) The condition for a line $lx + my + n = 0$ to be tangent to a circle $(x - h)^2 + (y - k)^2 = a^2$ is that the perpendicular distance from the center $(h, k)$ to the line must be equal to the radius $a$.
The perpendicular distance $d$ is given by:
$d = \frac{|lh + mk + n|}{\sqrt{l^2 + m^2}}$
Setting $d = a$:
$a = \frac{|lh + mk + n|}{\sqrt{l^2 + m^2}}$
Squaring both sides:
$a^2 = \frac{(lh + mk + n)^2}{l^2 + m^2}$
Therefore:
$(lh + mk + n)^2 = a^2(l^2 + m^2)$

(C) The equation of the circle is $(x - a)^2 + (y - b)^2 = r^2$,which has center $(a, b)$ and radius $r$.
The perpendicular distance $d$ from the center $(a, b)$ to the line $(x - a)\cos \alpha + (y - b)\sin \alpha - r = 0$ is given by:
$d = \frac{|(a - a)\cos \alpha + (b - b)\sin \alpha - r|}{\sqrt{\cos^2 \alpha + \sin^2 \alpha}}$
$d = \frac{|0 + 0 - r|}{\sqrt{1}} = |-r| = r$.
Since the perpendicular distance from the center to the line is equal to the radius $(d = r)$,the line is a tangent to the circle for all values of $\alpha$.

(B) The equation of the pair of tangents from a point $(x_1, y_1)$ to a circle $S = 0$ is given by $SS_1 = T^2$.
Here,the point is $(0, 0)$,so $S_1 = 0^2 + 0^2 - 2r(0) - 2h(0) + h^2 = h^2$.
The equation of the tangent $T$ at $(0, 0)$ is $x(0) + y(0) - r(x + 0) - h(y + 0) + h^2 = 0$,which simplifies to $T = -rx - hy + h^2$.
Substituting these into $SS_1 = T^2$:
$h^2(x^2 + y^2 - 2rx - 2hy + h^2) = (-rx - hy + h^2)^2$.
Expanding both sides:
$h^2x^2 + h^2y^2 - 2rh^2x - 2h^3y + h^4 = r^2x^2 + h^2y^2 + h^4 + 2rhxy - 2rh^2x - 2h^3y$.
Canceling common terms $h^2y^2, h^4, -2rh^2x, -2h^3y$:
$h^2x^2 = r^2x^2 + 2rhxy$.
$(h^2 - r^2)x^2 - 2rhxy = 0$.
$x((h^2 - r^2)x - 2rhy) = 0$.
Thus,the equations are $x = 0$ and $(h^2 - r^2)x - 2rhy = 0$.

(A) The equation of any line parallel to $\sqrt{3}x + y + 3 = 0$ is of the form $\sqrt{3}x + y + k = 0$.
Since this line is a tangent to the circle $x^2 + y^2 = a^2$,the perpendicular distance from the center $(0, 0)$ to the line must be equal to the radius $a$.
Using the formula for the distance from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$,which is $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$,we get:
$\frac{|\sqrt{3}(0) + 1(0) + k|}{\sqrt{(\sqrt{3})^2 + 1^2}} = a$
$\frac{|k|}{\sqrt{3 + 1}} = a$
$\frac{|k|}{2} = a$
$|k| = 2a$
$k = \pm 2a$
Substituting the value of $k$ back into the equation,we get $\sqrt{3}x + y \pm 2a = 0$.

(D) The equation of the circle is $x^2 + y^2 = 169$,which has center $(0, 0)$ and radius $r = 13$.
The equation of the tangent to the circle $x^2 + y^2 = r^2$ at point $(x_1, y_1)$ is given by $xx_1 + yy_1 = r^2$.
For the point $(5, 12)$,the tangent equation is $5x + 12y = 169$. The slope $m_1 = -\frac{5}{12}$.
For the point $(12, -5)$,the tangent equation is $12x - 5y = 169$. The slope $m_2 = \frac{12}{5}$.
Since $m_1 \times m_2 = (-\frac{5}{12}) \times (\frac{12}{5}) = -1$,the tangents are perpendicular to each other.
Therefore,the angle between the tangents is $90^o$.

(C) The equation of the circle is $x^2 + y^2 = 9$,which represents a circle centered at $(0, 0)$ with radius $r = \sqrt{9} = 3$.
For a line $x = k$ to be tangent to the circle,the perpendicular distance from the center $(0, 0)$ to the line $x - k = 0$ must be equal to the radius $r$.
The perpendicular distance $d$ is given by $d = \frac{|0 - k|}{\sqrt{1^2 + 0^2}} = |k|$.
Setting $d = r$,we get $|k| = 3$,which implies $k = 3$ or $k = -3$.
Therefore,the correct value of $k$ is $3$ or $-3$.

(B) The given line is $y \cos \alpha = x \sin \alpha + a \cos \alpha$.
Dividing by $\cos \alpha$ (assuming $\cos \alpha \neq 0$),we get $y = x \tan \alpha + a$.
This is in the form $y = mx + c$,where $m = \tan \alpha$ and $c = a$.
The condition for the line $y = mx + c$ to be a tangent to the circle $x^2 + y^2 = a^2$ is $c^2 = a^2(1 + m^2)$.
Substituting the values,we get $a^2 = a^2(1 + \tan^2 \alpha)$.
Since $a^2 \neq 0$,we have $1 = 1 + \tan^2 \alpha$,which implies $\tan^2 \alpha = 0$.
Alternatively,using the perpendicular distance from the center $(0, 0)$ to the line $x \sin \alpha - y \cos \alpha + a \cos \alpha = 0$ being equal to the radius $a$:
$\frac{|0 - 0 + a \cos \alpha|}{\sqrt{\sin^2 \alpha + \cos^2 \alpha}} = a$
$|a \cos \alpha| = a$
$|\cos \alpha| = 1$
$\cos^2 \alpha = 1$.

(A) Let $O$ be the center of the circle and $C$ be an external point. Let $CA$ and $CB$ be the two tangents drawn from $C$ to the circle touching at points $A$ and $B$ respectively.
In $\triangle OAC$ and $\triangle OBC$:
$1$. $OA = OB$ (Radii of the same circle)
$2$. $OC = OC$ (Common side)
$3$. $\angle OAC = \angle OBC = 90^o$ (Tangent is perpendicular to the radius at the point of contact)
By $RHS$ congruence criterion,$\triangle OAC \cong \triangle OBC$.
Therefore,by $CPCT$,$CA = CB$.
Thus,the two tangents drawn from an external point to a circle are always equal in length.

(A) The given circle equation is $x^2 + y^2 - 8x - 2y + 12 = 0$. The center of the circle is $(4, 1)$.
Substitute $y = -1$ into the circle equation to find the abscissa:
$x^2 + (-1)^2 - 8x - 2(-1) + 12 = 0$
$x^2 - 8x + 1 + 2 + 12 = 0$
$x^2 - 8x + 15 = 0$
$(x - 3)(x - 5) = 0$
So,$x = 3$ or $x = 5$. The points are $(3, -1)$ and $(5, -1)$.
The normal to a circle at any point always passes through the center $(4, 1)$.
For point $(3, -1)$,the slope of the normal is $m = \frac{1 - (-1)}{4 - 3} = \frac{2}{1} = 2$.
The equation is $y - (-1) = 2(x - 3)$ $\Rightarrow y + 1 = 2x - 6$ $\Rightarrow 2x - y - 7 = 0$.
For point $(5, -1)$,the slope of the normal is $m = \frac{1 - (-1)}{4 - 5} = \frac{2}{-1} = -2$.
The equation is $y - (-1) = -2(x - 5)$ $\Rightarrow y + 1 = -2x + 10$ $\Rightarrow 2x + y - 9 = 0$.

(A) The point of contact must satisfy both the equation of the line and the equation of the circle.
Checking option $A$: $(1, -2)$
For the line: $1 - (-2) - 3 = 1 + 2 - 3 = 0$. (Satisfied)
For the circle: $(1)^2 + (-2)^2 - 4(1) + 6(-2) + 11 = 1 + 4 - 4 - 12 + 11 = 0$. (Satisfied)
Since the point $(1, -2)$ satisfies both equations,it is the point of contact.

(B) line is a normal to a circle if and only if it passes through the centre of the circle.
Given the circle has a centre at $(a, b)$ and the line equation is $y = mx + c$.
Substituting the coordinates of the centre $(a, b)$ into the equation of the line,we get:
$b = m(a) + c$
$b = ma + c$

(C) The equation of the circle is $x^2 + y^2 + 4x + 6y - 39 = 0$. The center of the circle is $(-g, -f) = (-2, -3)$.
The normal to a circle at any point always passes through the center of the circle.
The normal passes through the point $(2, 3)$ and the center $(-2, -3)$.
The slope of the normal is $m = \frac{3 - (-3)}{2 - (-2)} = \frac{6}{4} = \frac{3}{2}$.
The equation of the normal line is $y - 3 = \frac{3}{2}(x - 2)$,which simplifies to $2y - 6 = 3x - 6$,or $3x - 2y = 0$.
To find the other point of intersection,substitute $x = \frac{2y}{3}$ into the circle equation:
$(\frac{2y}{3})^2 + y^2 + 4(\frac{2y}{3}) + 6y - 39 = 0$
$\frac{4y^2}{9} + y^2 + \frac{8y}{3} + 6y - 39 = 0$
Multiply by $9$: $4y^2 + 9y^2 + 24y + 54y - 351 = 0$
$13y^2 + 78y - 351 = 0$
Divide by $13$: $y^2 + 6y - 27 = 0$
$(y - 3)(y + 9) = 0$,so $y = 3$ or $y = -9$.
If $y = 3$,$x = \frac{2(3)}{3} = 2$ (this is the given point).
If $y = -9$,$x = \frac{2(-9)}{3} = -6$.
Thus,the other point is $(-6, -9)$.

(C) The given circle is $x^2 + y^2 - 2x = 0$.
Comparing this with the standard form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -1$ and $f = 0$.
The centre of the circle is $(-g, -f) = (1, 0)$.
Any line parallel to $x + 2y = 3$ is of the form $x + 2y + \lambda = 0$.
Since the normal to a circle always passes through its centre,the line $x + 2y + \lambda = 0$ must pass through $(1, 0)$.
Substituting the point $(1, 0)$ into the equation: $1 + 2(0) + \lambda = 0$,which gives $\lambda = -1$.
Thus,the equation of the normal is $x + 2y - 1 = 0$.

(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 $xx_1 + yy_1 = r^2$.
Here,$x_1 = \frac{ab^2}{a^2 + b^2}$,$y_1 = \frac{a^2b}{a^2 + b^2}$,and $r^2 = \frac{a^2b^2}{a^2 + b^2}$.
Substituting these values into the formula:
$x \left( \frac{ab^2}{a^2 + b^2} \right) + y \left( \frac{a^2b}{a^2 + b^2} \right) = \frac{a^2b^2}{a^2 + b^2}$
Dividing both sides by $\frac{ab}{a^2 + b^2}$ (assuming $a, b \neq 0$):
$xb + ya = ab$
Dividing by $ab$:
$\frac{xb}{ab} + \frac{ya}{ab} = \frac{ab}{ab}$
$\frac{x}{a} + \frac{y}{b} = 1$.

(A) The equation of the line passing through the origin is $y = mx$,which can be written as $mx - y = 0$.
The circle is given by $(x - 4)^2 + (y + 5)^2 = 25$,which has center $(4, -5)$ and radius $r = 5$.
For the line to be tangent to the circle,the perpendicular distance from the center $(4, -5)$ to the line $mx - y = 0$ must be equal to the radius $r = 5$.
Using the distance formula: $\frac{|m(4) - (-5)|}{\sqrt{m^2 + (-1)^2}} = 5$.
$|4m + 5| = 5\sqrt{m^2 + 1}$.
Squaring both sides: $(4m + 5)^2 = 25(m^2 + 1)$.
$16m^2 + 40m + 25 = 25m^2 + 25$.
$9m^2 - 40m = 0$.
$m(9m - 40) = 0$.
Thus,$m = 0$ or $m = \frac{40}{9}$.

(A) The equation of the pair of tangents from the origin $(0, 0)$ to the circle ${x^2} + {y^2} + 2gx + 2fy + c = 0$ is given by $S S_1 = T^2$.
Here,$S = {x^2} + {y^2} + 2gx + 2fy + c$,$S_1 = c$,and $T = gx + fy + c$.
So,$c({x^2} + {y^2} + 2gx + 2fy + c) = (gx + fy + c)^2$.
Expanding this,we get $c{x^2} + c{y^2} + 2gcx + 2fcy + {c^2} = {g^2}{x^2} + {f^2}{y^2} + {c^2} + 2gfxy + 2gcx + 2fcy$.
Simplifying,we get $(c - {g^2}){x^2} - 2gfxy + (c - {f^2}){y^2} = 0$.
Since the tangents are perpendicular,the sum of the coefficients of ${x^2}$ and ${y^2}$ must be zero.
Therefore,$(c - {g^2}) + (c - {f^2}) = 0$.
This implies ${g^2} + {f^2} = 2c$.

(D) Let the circle be $S \equiv x^2 + y^2 + 2gx + 2fy + c = 0$. The origin is $O(0, 0)$.
The tangents $OP$ and $OQ$ are drawn from the origin to the circle.
The points of contact $P$ and $Q$ lie on the circle,and the line $PQ$ is the chord of contact.
The equation of the chord of contact $PQ$ is $T = 0$,which is $gx + fy + c = 0$.
The circle passing through the points of intersection of the circle $S=0$ and the line $PQ=0$ is given by $S + \lambda(gx + fy + c) = 0$.
Since the circle passes through the origin $(0, 0)$,we have $c + \lambda(c) = 0$,which implies $\lambda = -1$.
Substituting $\lambda = -1$,the equation of the circle passing through $O, P, Q$ is $x^2 + y^2 + 2gx + 2fy + c - (gx + fy + c) = 0$,which simplifies to $x^2 + y^2 + gx + fy = 0$.
The circumcentre of $\triangle OPQ$ is the centre of this circle,which is $\left(-\frac{g}{2}, -\frac{f}{2}\right)$.
Since this value is not among the given options,the correct answer is $(d)$.

(C) To find the point of contact,substitute the equation of the line $x = 0$ into the equation of the circle $x^2 + y^2 - 2x - 6y + 9 = 0$.
Substituting $x = 0$ gives:
$(0)^2 + y^2 - 2(0) - 6y + 9 = 0$
$y^2 - 6y + 9 = 0$
This can be written as $(y - 3)^2 = 0$.
Solving for $y$,we get $y = 3$.
Therefore,the point of contact is $(0, 3)$.

(A) The equation of the circle is ${x^2} + {y^2} - 4y = 0$.
Rewriting it as ${x^2} + {(y - 2)^2} = 4$,we identify the center as $(0, 2)$ and the radius $r = 2$.
For the line $mx - y + c = 0$ to be tangent to the circle,the perpendicular distance from the center $(0, 2)$ to the line must equal the radius $r$.
$\frac{|m(0) - 2 + c|}{\sqrt{m^2 + (-1)^2}} = 2$
$|c - 2| = 2\sqrt{1 + m^2}$
$c - 2 = \pm 2\sqrt{1 + m^2}$
$c = 2 \pm 2\sqrt{1 + m^2} = 2(1 \pm \sqrt{1 + m^2})$.

(A) The equation of the circle is $x^2 + y^2 = 4$. The point of contact is $P(1, \sqrt{3})$.
The equation of the tangent at $(x_1, y_1)$ is $xx_1 + yy_1 = r^2$. Substituting the values,we get $x(1) + y(\sqrt{3}) = 4$,which simplifies to $x + \sqrt{3}y = 4$.
The tangent intersects the $x$-axis at $y = 0$,so $x = 4$. The point is $A(4, 0)$.
The normal at $(1, \sqrt{3})$ passes through the center $(0, 0)$ and the point $(1, \sqrt{3})$. Its equation is $y = \sqrt{3}x$.
The normal intersects the $x$-axis at the origin $O(0, 0)$.
The triangle is formed by the vertices $O(0, 0)$,$P(1, \sqrt{3})$,and $A(4, 0)$.
The area of the triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Area $= \frac{1}{2} |0(\sqrt{3} - 0) + 1(0 - 0) + 4(0 - \sqrt{3})| = \frac{1}{2} |-4\sqrt{3}| = 2\sqrt{3}$ square units.

(D) The equation of the line is $x - y + a\sqrt{2} = 0$.
The equation of the tangent to the circle $x^2 + y^2 = a^2$ at point $(h, k)$ is $hx + ky = a^2$.
Comparing the given line $x - y = -a\sqrt{2}$ with $hx + ky = a^2$,we have:
$\frac{h}{1} = \frac{k}{-1} = \frac{a^2}{-a\sqrt{2}}$.
From this,$h = \frac{a^2}{-a\sqrt{2}} = -\frac{a}{\sqrt{2}}$ and $k = \frac{a^2}{a\sqrt{2}} = \frac{a}{\sqrt{2}}$.
Thus,the point of contact is $\left( -\frac{a}{\sqrt{2}}, \frac{a}{\sqrt{2}} \right)$.

(B) The equation of the tangent to the circle $x^2 + y^2 = 5$ at the point $(x_1, y_1) = (1, -2)$ is given by $xx_1 + yy_1 = r^2$.
Substituting the values,we get $x(1) + y(-2) = 5$,which simplifies to $x - 2y = 5$.
To find the point of contact that also touches the circle $x^2 + y^2 - 8x + 6y + 20 = 0$,we check which of the given options satisfies the equation of the tangent $x - 2y = 5$.
For option $(B)$,$(3, -1)$: $3 - 2(-1) = 3 + 2 = 5$. This satisfies the equation.
Thus,the point of contact is $(3, -1)$.

(A) The equation of the circle is $x^2 + y^2 - 3x - 6y - 10 = 0$. Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -\frac{3}{2}$ and $f = -3$.
The center of the circle is $(-g, -f) = (\frac{3}{2}, 3)$.
The normal to a circle at any point $(x_1, y_1)$ always passes through the center of the circle.
The slope of the line passing through the center $(\frac{3}{2}, 3)$ and the point $(-3, 4)$ is $m = \frac{4 - 3}{-3 - \frac{3}{2}} = \frac{1}{-\frac{9}{2}} = -\frac{2}{9}$.
The equation of the normal line passing through $(-3, 4)$ with slope $m = -\frac{2}{9}$ is given by $y - y_1 = m(x - x_1)$.
$y - 4 = -\frac{2}{9}(x + 3)$
$9(y - 4) = -2(x + 3)$
$9y - 36 = -2x - 6$
$2x + 9y - 30 = 0$.

(C) The line $y = x + c$ intersects the circle $x^2 + y^2 = 1$ at two coincident points,which means the line is a tangent to the circle.
For a line $y = mx + c$ to be a tangent to the circle $x^2 + y^2 = r^2$,the condition is $c^2 = r^2(1 + m^2)$.
Here,$m = 1$ and $r^2 = 1$.
Substituting these values,we get $c^2 = 1(1 + 1^2) = 2$.
Therefore,$c = \pm \sqrt{2}$.

(B) The equation of the circle is $x^2 + y^2 = 25$,which is of the form $x^2 + y^2 = a^2$,where $a = 5$.
The condition for the line $y = mx + c$ to be a tangent to the circle $x^2 + y^2 = a^2$ is $c = \pm a\sqrt{1 + m^2}$.
Substituting $a = 5$,we get $c = \pm 5\sqrt{1 + m^2}$.
Thus,the equation of the tangent is $y = mx \pm 5\sqrt{1 + m^2}$.
Comparing this with the given options,the correct line is $y = mx + 5\sqrt{1 + m^2}$.

(B) The equation of the circle is $ax^2 + ay^2 = r^2$.
Dividing by $a$,we get the standard form: $x^2 + y^2 - \frac{r^2}{a} = 0$.
The square of the length of the tangent from a point $(\alpha, \beta)$ to a circle $S = x^2 + y^2 + 2gx + 2fy + c = 0$ is given by $S_1 = \alpha^2 + \beta^2 + 2g\alpha + 2f\beta + c$.
Substituting the values from our equation,we get $S_1 = \alpha^2 + \beta^2 - \frac{r^2}{a}$.

(C) The normal to a circle at any point always passes through the center of the circle.
Given the normal equation $y - x + 3 = 0$,we can rewrite it as $y = x - 3$.
For a circle $(x - h)^2 + (y - k)^2 = r^2$,the center is $(h, k)$.
Testing the centers of the given options:
For option $(c)$,the circle is $(x - 3)^2 + y^2 = 9$,which has the center $(3, 0)$.
Substituting the center $(3, 0)$ into the normal equation $y - x + 3 = 0$:
$0 - 3 + 3 = 0$.
Since the center satisfies the equation,the normal passes through the center.
Thus,option $(c)$ is the correct circle.

(C) The equation of the circle is $x^2 + y^2 - 2x - 4y + 3 = 0$.
The equation of the tangent to the circle $x^2 + y^2 + 2gx + 2fy + c' = 0$ at the point $(x_1, y_1)$ is given by $xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c' = 0$.
Comparing the given circle with the standard form,we have $g = -1$,$f = -2$,and $c' = 3$.
Substituting the point $(2, 3)$ into the tangent formula:
$x(2) + y(3) - 1(x + 2) - 2(y + 3) + 3 = 0$
$2x + 3y - x - 2 - 2y - 6 + 3 = 0$
$x + y - 5 = 0$
Rearranging into the slope-intercept form $y = mx + c$:
$y = -x + 5$
Comparing this with $y = mx + c$,we get $c = 5$.

(A) Let the circle be $S: x^2 + y^2 + 2gx + 2fy + c = 0$.
The length of the tangent from a point $P(x_1, y_1)$ to the circle is defined as $\sqrt{S_1}$,where $S_1$ is the value obtained by substituting the coordinates of point $P$ into the equation of the circle.
Thus,the length of the tangent is $\sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c}$.

(C) The equation of the tangent to the circle $x^2 + y^2 = r^2$ at a point $(x_1, y_1)$ on the circle is given by $xx_1 + yy_1 = r^2$.
Given the point is $(a, b)$,the equation of the tangent is $ax + by = r^2$.
Rewriting this in the form $ax + by - r^2 = 0$,we compare it with the given equation $ax + by - \lambda = 0$.
Thus,$\lambda = r^2$.

(B) The centre of the circle is $C(-6, 8)$.
Since the circle passes through the origin $O(0, 0)$,the radius $OC$ is the line segment connecting $(0, 0)$ and $(-6, 8)$.
The slope of the radius $OC$ is $m_{radius} = \frac{8 - 0}{-6 - 0} = \frac{8}{-6} = -\frac{4}{3}$.
The tangent at the origin is perpendicular to the radius $OC$.
Therefore,the slope of the tangent $m_{tangent} = -\frac{1}{m_{radius}} = -\frac{1}{-4/3} = \frac{3}{4}$.
The equation of the tangent passing through the origin $(0, 0)$ with slope $m = \frac{3}{4}$ is $y - 0 = \frac{3}{4}(x - 0)$,which simplifies to $4y = 3x$ or $3x - 4y = 0$.

(B) The center of the circle $C$ is $(-g, -f)$ and the radius $r$ is $\sqrt{g^2 + f^2 - c}$.
Let $O$ be the origin $(0, 0)$. The length of the tangent $OA$ from the origin to the circle is given by $\sqrt{S_1}$,where $S_1$ is the value of the circle equation at $(0, 0)$.
Thus,$OA = \sqrt{0^2 + 0^2 + 2g(0) + 2f(0) + c} = \sqrt{c}$.
In the right-angled triangle $\Delta OAC$,$\angle OAC = 90^\circ$ because the tangent is perpendicular to the radius at the point of contact.
The area of $\Delta OAC = \frac{1}{2} \times OA \times AC = \frac{1}{2} \times \sqrt{c} \times \sqrt{g^2 + f^2 - c}$.
The quadrilateral $OACB$ is composed of two congruent triangles,$\Delta OAC$ and $\Delta OBC$.
Therefore,the area of quadrilateral $OACB = 2 \times \text{Area}(\Delta OAC) = 2 \times \frac{1}{2} \times \sqrt{c} \times \sqrt{g^2 + f^2 - c} = \sqrt{c(g^2 + f^2 - c)}$.

(D) The equation of the tangent at point $(a \cos \alpha, a \sin \alpha)$ to the circle $x^2 + y^2 = a^2$ is given by $x(a \cos \alpha) + y(a \sin \alpha) = a^2$.
Rearranging this into the slope-intercept form $y = mx + c$,we get $y = -\frac{\cos \alpha}{\sin \alpha} x + \frac{a}{\sin \alpha}$.
Thus,the gradient (slope) $m$ is $- \frac{\cos \alpha}{\sin \alpha} = - \cot \alpha$.

(C) The equation of any circle passing through $(0, a)$ and $(0, -a)$ is of the form $x^2 + y^2 + \lambda x - a^2 = 0$.
This circle touches the line $mx - y + c = 0$. The perpendicular distance from the center $(-\lambda/2, 0)$ to the line is equal to the radius $r = \sqrt{(\lambda/2)^2 + a^2}$.
Thus,$\frac{|m(-\lambda/2) - 0 + c|}{\sqrt{m^2 + 1}} = \sqrt{\frac{\lambda^2}{4} + a^2}$.
Squaring both sides: $\frac{(c - m\lambda/2)^2}{m^2 + 1} = \frac{\lambda^2}{4} + a^2$.
$(c - m\lambda/2)^2 = (m^2 + 1)(\frac{\lambda^2}{4} + a^2)$.
$c^2 - mc\lambda + \frac{m^2\lambda^2}{4} = \frac{m^2\lambda^2}{4} + m^2a^2 + \frac{\lambda^2}{4} + a^2$.
$\frac{\lambda^2}{4} + mc\lambda + (a^2(m^2 + 1) - c^2) = 0$.
This is a quadratic in $\lambda$,giving two circles with parameters $\lambda_1$ and $\lambda_2$. For these circles to be orthogonal,$2g_1g_2 + 2f_1f_2 = c_1 + c_2$.
Here,$g_1 = \lambda_1/2, g_2 = \lambda_2/2, f_1 = 0, f_2 = 0, c_1 = -a^2, c_2 = -a^2$.
So,$2(\frac{\lambda_1}{2})(\frac{\lambda_2}{2}) = -a^2 - a^2$ $\Rightarrow \frac{\lambda_1\lambda_2}{2} = -2a^2$ $\Rightarrow \lambda_1\lambda_2 = -4a^2$.
From the quadratic equation,the product of roots $\lambda_1\lambda_2 = 4(a^2(m^2 + 1) - c^2)$.
Equating the two: $4(a^2(m^2 + 1) - c^2) = -4a^2$.
$a^2(m^2 + 1) - c^2 = -a^2$.
$c^2 = a^2(m^2 + 1 + 1) = a^2(m^2 + 2)$.

(B) The given circle is $x^2 + y^2 + 4x - 4y + 4 = 0$.
Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = 2$,$f = -2$,and $c = 4$.
The center is $(-g, -f) = (-2, 2)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{4 + 4 - 4} = 2$.
$A$ line making equal intercepts on the positive coordinate axes has the form $x + y = a$,where $a > 0$,or $x + y - a = 0$.
Since this line is a tangent to the circle,the perpendicular distance from the center $(-2, 2)$ to the line must equal the radius $r = 2$.
Thus,$\left| \frac{-2 + 2 - a}{\sqrt{1^2 + 1^2}} \right| = 2$.
$\left| \frac{-a}{\sqrt{2}} \right| = 2 \Rightarrow |-a| = 2\sqrt{2}$.
Since the intercepts are positive,$a = 2\sqrt{2}$.
Therefore,the equation of the tangent is $x + y = 2\sqrt{2}$.

(C) Let $P = (\alpha, \beta)$ be the external point and $C$ be the center of the circle $(0, 0)$. Let the tangents from $P$ touch the circle at $T_1$ and $T_2$.
In the right-angled triangle $\triangle PT_1C$,the angle $\angle CPT_1 = \theta$ (where $2\theta$ is the total angle between the tangents).
We have $CT_1 = a$ (radius) and $PT_1 = \sqrt{OP^2 - CT_1^2} = \sqrt{\alpha^2 + \beta^2 - a^2}$.
Thus,$\tan \theta = \frac{CT_1}{PT_1} = \frac{a}{\sqrt{\alpha^2 + \beta^2 - a^2}}$.
Therefore,$\theta = \tan^{-1}\left(\frac{a}{\sqrt{\alpha^2 + \beta^2 - a^2}}\right)$.
The total angle between the tangents is $2\theta = 2\tan^{-1}\left(\frac{a}{\sqrt{\alpha^2 + \beta^2 - a^2}}\right)$.

(D) The given circle is $x^2 + y^2 - 2x - 4y - 4 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -1$,$f = -2$,and $c = -4$. The center is $(1, 2)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{1 + 4 + 4} = 3$.
The line perpendicular to $3x - 4y - 1 = 0$ is of the form $4x + 3y + k = 0$.
The perpendicular distance from the center $(1, 2)$ to the tangent $4x + 3y + k = 0$ must be equal to the radius $r = 3$.
Using the distance formula: $\frac{|4(1) + 3(2) + k|}{\sqrt{4^2 + 3^2}} = 3$.
$\frac{|4 + 6 + k|}{5} = 3 \implies |10 + k| = 15$.
This gives $10 + k = 15$ or $10 + k = -15$.
So,$k = 5$ or $k = -25$.
The possible tangents are $4x + 3y + 5 = 0$ and $4x + 3y - 25 = 0$. Comparing with the options,the correct answer is $4x + 3y - 25 = 0$.

(B) Let the equation of the tangent be $\frac{x}{x_1} + \frac{y}{y_1} = 1$. The area of the triangle formed by this line with the coordinate axes is $\frac{1}{2} |x_1 y_1| = a^2$,so $|x_1 y_1| = 2a^2$.
The line equation is $y_1 x + x_1 y - x_1 y_1 = 0$.
Since this line is tangent to the circle $x^2 + y^2 = a^2$,the perpendicular distance from the origin $(0, 0)$ to the line must be equal to the radius $a$:
$\frac{|-x_1 y_1|}{\sqrt{y_1^2 + x_1^2}} = a$
Squaring both sides:
$\frac{x_1^2 y_1^2}{x_1^2 + y_1^2} = a^2$
Substitute $x_1^2 y_1^2 = (2a^2)^2 = 4a^4$:
$\frac{4a^4}{x_1^2 + y_1^2} = a^2 \implies x_1^2 + y_1^2 = 4a^2$.
We have $|x_1 y_1| = 2a^2$ and $x_1^2 + y_1^2 = 4a^2$.
$(|x_1| + |y_1|)^2 = x_1^2 + y_1^2 + 2|x_1 y_1| = 4a^2 + 4a^2 = 8a^2 \implies |x_1| + |y_1| = 2a\sqrt{2}$.
$(|x_1| - |y_1|)^2 = x_1^2 + y_1^2 - 2|x_1 y_1| = 4a^2 - 4a^2 = 0 \implies |x_1| = |y_1|$.
Thus,$|x_1| = |y_1| = a\sqrt{2}$.
Substituting these into the intercept form $\frac{x}{x_1} + \frac{y}{y_1} = 1$,we get $\frac{x}{\pm a\sqrt{2}} + \frac{y}{\pm a\sqrt{2}} = 1$,which simplifies to $x \pm y = \pm a\sqrt{2}$.