Tangent and normal to a circle Questions in English

(B) The given circle is $x^2 + y^2 - 4x - 8y - 5 = 0$. Comparing this with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -2$,$f = -4$,and $c = -5$.
The center of the circle is $(-g, -f) = (2, 4)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{2^2 + 4^2 - (-5)} = \sqrt{4 + 16 + 5} = \sqrt{25} = 5$.
Since the line $3x - 4y - \lambda = 0$ touches the circle,the perpendicular distance from the center $(2, 4)$ to the line must be equal to the radius $r$.
Distance $d = \frac{|3(2) - 4(4) - \lambda|}{\sqrt{3^2 + (-4)^2}} = 5$.
$\frac{|6 - 16 - \lambda|}{\sqrt{9 + 16}} = 5$.
$\frac{|-10 - \lambda|}{5} = 5$.
$|-10 - \lambda| = 25$.
This implies $-10 - \lambda = 25$ or $-10 - \lambda = -25$.
If $-10 - \lambda = 25$,then $\lambda = -35$.
If $-10 - \lambda = -25$,then $\lambda = 15$.
Thus,$\lambda = -35, 15$.

(C) The equation of the circle is $x^2 + y^2 - 2ax - 2by + b^2 = 0$.
Comparing this with the general equation $x^2 + y^2 + 2gx + 2fy + c = 0$,we get the center $(g, f) = (a, b)$ and radius $r = \sqrt{a^2 + b^2 - b^2} = \sqrt{a^2} = |a|$.
For tangents drawn from the origin $(0, 0)$ to be perpendicular,the distance from the center $(a, b)$ to the origin must be equal to $\sqrt{2} \times r$.
Distance from $(0, 0)$ to $(a, b)$ is $\sqrt{a^2 + b^2}$.
So,$\sqrt{a^2 + b^2} = \sqrt{2} \times |a|$.
Squaring both sides,we get $a^2 + b^2 = 2a^2$,which simplifies to $b^2 = a^2$ or $a^2 = b^2$.

(A) line is normal to a circle if and only if it passes through the center of the circle.
The center of the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is $(-g, -f)$.
Since the line $lx + my + n = 0$ is normal to the circle,the point $(-g, -f)$ must satisfy the equation of the line.
Substituting $x = -g$ and $y = -f$ into the line equation:
$l(-g) + m(-f) + n = 0$
$-lg - mf + n = 0$
Multiplying by $-1$,we get:
$lg + mf - n = 0$.

(B) The given equations of the tangents are $2x - 4y - 9 = 0$ and $6x - 12y + 7 = 0$.
To make the coefficients of $x$ and $y$ the same,multiply the first equation by $3$: $6x - 12y - 27 = 0$.
Since the tangents are parallel,the distance between them is equal to the diameter $d$ of the circle.
The distance $d$ between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.
Here,$A = 6, B = -12, C_1 = -27, C_2 = 7$.
$d = \frac{|-27 - 7|}{\sqrt{6^2 + (-12)^2}} = \frac{|-34|}{\sqrt{36 + 144}} = \frac{34}{\sqrt{180}} = \frac{34}{6\sqrt{5}} = \frac{17}{3\sqrt{5}}$.
The radius $r$ is half of the diameter: $r = \frac{d}{2} = \frac{17}{2 \times 3\sqrt{5}} = \frac{17}{6\sqrt{5}}$.

(B) The equation of the circle is $x^2 + y^2 = a^2$,which has center $(0, 0)$ and radius $a$.
Any line parallel to $y = mx + c$ is of the form $y = mx + k$.
For this line to be a tangent to the circle,the perpendicular distance from the center $(0, 0)$ to the line $mx - y + k = 0$ must be equal to the radius $a$.
The distance formula is $d = \frac{|m(0) - (0) + k|}{\sqrt{m^2 + (-1)^2}} = a$.
This simplifies to $\frac{|k|}{\sqrt{m^2 + 1}} = a$,which gives $|k| = a\sqrt{1 + m^2}$.
Thus,$k = \pm a\sqrt{1 + m^2}$.
Substituting $k$ back into the line equation,we get $y = mx \pm a\sqrt{1 + m^2}$.

(A) Given the equation of the circle is ${x^2} + {y^2} - 4x - 6y - 12 = 0$.
To find the point of contact with the line $x = 7$,substitute $x = 7$ into the circle's equation:
${(7)^2} + {y^2} - 4(7) - 6y - 12 = 0$
$49 + {y^2} - 28 - 6y - 12 = 0$
${y^2} - 6y + 9 = 0$
This is a perfect square trinomial:
${(y - 3)^2} = 0$
$y = 3$
Thus,the point of contact is $(7, 3)$.

(A) The equation of any tangent to the circle $x^2 + y^2 = b^2$ is given by $y = mx \pm b\sqrt{1 + m^2}$.
Since the given tangent is $y = mx - b\sqrt{1 + m^2}$,it is a tangent to the first circle.
For this line to be a tangent to the second circle $(x - a)^2 + y^2 = b^2$,the perpendicular distance from the center $(a, 0)$ to the line $mx - y - b\sqrt{1 + m^2} = 0$ must be equal to the radius $b$.
Thus,$\frac{|ma - 0 - b\sqrt{1 + m^2}|}{\sqrt{m^2 + 1}} = b$.
Since $a > 2b$,we have $ma > b\sqrt{1 + m^2}$,so $ma - b\sqrt{1 + m^2} = b\sqrt{m^2 + 1}$.
$ma = 2b\sqrt{1 + m^2}$.
Squaring both sides,$m^2 a^2 = 4b^2(1 + m^2) = 4b^2 + 4b^2 m^2$.
$m^2(a^2 - 4b^2) = 4b^2$.
$m^2 = \frac{4b^2}{a^2 - 4b^2}$.
Since we need the positive value of $m$,$m = \frac{2b}{\sqrt{a^2 - 4b^2}}$.

(B) Let the point of contact be $P(x_1, y_1)$.
Since $P$ lies on the line $x + 2y + 12 = 0$,we have $x_1 + 2y_1 = -12$ $(i)$.
The line $x + 2y + 12 = 0$ has a slope $m_2 = -\frac{1}{2}$.
The radius $OP$ is perpendicular to the tangent line,so the slope of $OP$ is $m_1 = -\frac{1}{m_2} = 2$.
The slope of $OP$ is also given by $\frac{y_1 - 1}{x_1 + 1}$.
Equating the slopes: $\frac{y_1 - 1}{x_1 + 1} = 2$ $\Rightarrow y_1 - 1 = 2x_1 + 2$ $\Rightarrow 2x_1 - y_1 = -3$ $(ii)$.
Solving equations $(i)$ and $(ii)$:
From $(ii)$,$y_1 = 2x_1 + 3$.
Substituting into $(i)$: $x_1 + 2(2x_1 + 3) = -12$ $\Rightarrow x_1 + 4x_1 + 6 = -12$ $\Rightarrow 5x_1 = -18$ $\Rightarrow x_1 = -\frac{18}{5}$.
Then $y_1 = 2(-\frac{18}{5}) + 3 = -\frac{36}{5} + \frac{15}{5} = -\frac{21}{5}$.
Thus,the point of contact is $\left( -\frac{18}{5}, -\frac{21}{5} \right)$.

(C) The equation of the tangent to the circle $x^2 + y^2 = a^2$ at the point $(x_1, y_1)$ is given by $xx_1 + yy_1 = a^2$.
Substituting the point $(h, h)$,we get $hx + hy = a^2$,which can be rewritten as $hx + hy - a^2 = 0$.
This is in the form $Ax + By + C = 0$,where the slope $m = -\frac{A}{B}$.
Thus,the slope $m = -\frac{h}{h} = -1$ (provided $h \neq 0$).

(D) The given equation of the circle is $2(x^2 + y^2) = 5$,which can be rewritten as $x^2 + y^2 = \frac{5}{2}$.
Comparing this with the standard form $x^2 + y^2 = r^2$,the radius $r = \sqrt{\frac{5}{2}}$.
The center of the circle is $(0, 0)$.
$A$ line $ax + by + c = 0$ touches a circle with center $(h, k)$ and radius $r$ if the perpendicular distance from the center to the line equals the radius.
Thus,$\frac{|4(0) + 3(0) + \lambda|}{\sqrt{4^2 + 3^2}} = \sqrt{\frac{5}{2}}$.
$\frac{|\lambda|}{\sqrt{16 + 9}} = \sqrt{\frac{5}{2}}$.
$\frac{|\lambda|}{5} = \sqrt{\frac{5}{2}}$.
$|\lambda| = 5 \times \frac{\sqrt{5}}{\sqrt{2}} = \frac{5\sqrt{5}}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{5\sqrt{10}}{2}$.
Therefore,$\lambda = \pm \frac{5\sqrt{10}}{2}$.

(A) The given equation of the circle is $x^2 + y^2 + 2x + 4y + 3 = 0$.
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 $(-g, -f) = (-1, -2)$.
The normal to a circle at any point always passes through the center of the circle.
Thus,the normal passes through the points $(-1, -2)$ and $(-2, -3)$.
The gradient (slope) $m$ of the normal is given by the formula $m = \frac{y_2 - y_1}{x_2 - x_1}$.
Substituting the coordinates: $m = \frac{-3 - (-2)}{-2 - (-1)} = \frac{-3 + 2}{-2 + 1} = \frac{-1}{-1} = 1$.
Therefore,the gradient of the normal is $1$.

(B) The given line is $y = mx + c$. The slope of this line is $m$.
Any line perpendicular to this line will have a slope of $-1/m$.
Let the equation of the tangent be $y = -\frac{1}{m}x + k$,which can be rewritten as $x + my - mk = 0$.
The condition for a line $Ax + By + C = 0$ to be a tangent to the circle $x^2 + y^2 = a^2$ is that the perpendicular distance from the center $(0, 0)$ to the line equals the radius $a$.
Thus,$\frac{|-mk|}{\sqrt{1^2 + m^2}} = a$.
$|mk| = a\sqrt{1 + m^2}$,so $mk = \pm a\sqrt{1 + m^2}$.
Substituting $mk$ back into the equation $x + my = mk$,we get $x + my = \pm a\sqrt{1 + m^2}$.

(B) The center of the circle is $C = (a, b)$ and it passes through the origin $O = (0, 0)$.
The radius $OC$ connects the origin to the center. The slope of the radius $OC$ is $m_{radius} = \frac{b - 0}{a - 0} = \frac{b}{a}$.
The tangent at the origin is perpendicular to the radius $OC$ at the point of contact $(0, 0)$.
Therefore,the slope of the tangent $m_{tangent}$ is the negative reciprocal of the slope of the radius:
$m_{tangent} = -\frac{1}{m_{radius}} = -\frac{a}{b}$.
Using the point-slope form of a line passing through $(0, 0)$ with slope $m = -\frac{a}{b}$:
$y - 0 = -\frac{a}{b}(x - 0)$
$y = -\frac{a}{b}x$
$by = -ax$
$ax + by = 0$.

(A) 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 circle equation $x^2 + y^2 + 2x - 6y - 6 = 0$,we have $g = 1$,$f = -3$,and $c = -6$.
The point is $(4, 5)$,so $x_1 = 4$ and $y_1 = 5$.
Length of the tangent $= \sqrt{4^2 + 5^2 + 2(4) - 6(5) - 6}$
$= \sqrt{16 + 25 + 8 - 30 - 6}$
$= \sqrt{49 - 36}$
$= \sqrt{13}$.

(C) The given circle is $x^2 + y^2 = 4$,which has its center at $(0, 0)$ and radius $r = 2$.
Any line parallel to $x + 2y + 3 = 0$ can be written in the form $x + 2y + \lambda = 0$.
For this line to be a tangent to the circle,the perpendicular distance from the center $(0, 0)$ to the line must be equal to the radius $r = 2$.
The perpendicular distance $d$ from $(x_1, y_1)$ to $Ax + By + C = 0$ is given by $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.
Substituting the values,we get:
$\frac{|1(0) + 2(0) + \lambda|}{\sqrt{1^2 + 2^2}} = 2$
$\frac{|\lambda|}{\sqrt{5}} = 2$
$|\lambda| = 2\sqrt{5}$
$\lambda = \pm 2\sqrt{5}$
Substituting $\lambda$ back into the equation of the line,we get $x + 2y \pm 2\sqrt{5} = 0$,or $x + 2y = \pm 2\sqrt{5}$.

(C) The given equation of the circle is $2x^2 + 2y^2 - 2x - 5y + 3 = 0$.
Dividing by $2$,we get $x^2 + y^2 - x - \frac{5}{2}y + \frac{3}{2} = 0$.
The equation of the tangent at $(x_1, y_1)$ is $xx_1 + yy_1 - \frac{1}{2}(x + x_1) - \frac{5}{4}(y + y_1) + \frac{3}{2} = 0$.
Substituting $(1, 1)$,we get $x + y - \frac{1}{2}(x + 1) - \frac{5}{4}(y + 1) + \frac{3}{2} = 0$.
Multiplying by $4$,we get $4x + 4y - 2x - 2 - 5y - 5 + 6 = 0$,which simplifies to $2x - y - 1 = 0$.
The slope of the tangent is $m_t = 2$.
The slope of the normal is $m_n = -\frac{1}{m_t} = -\frac{1}{2}$.
The equation of the normal at $(1, 1)$ is $y - 1 = -\frac{1}{2}(x - 1)$.
$2y - 2 = -x + 1$,which gives $x + 2y = 3$.

(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 $\sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c}$.
Given the point $(3, -4)$ and the circle $x^2 + y^2 - 4x - 6y + 3 = 0$.
Let $L$ be the length of the tangent.
$L^2 = x_1^2 + y_1^2 - 4x_1 - 6y_1 + 3$
Substitute $x_1 = 3$ and $y_1 = -4$:
$L^2 = (3)^2 + (-4)^2 - 4(3) - 6(-4) + 3$
$L^2 = 9 + 16 - 12 + 24 + 3$
$L^2 = 25 - 12 + 24 + 3$
$L^2 = 13 + 24 + 3 = 40$.
Thus,the square of the length of the tangent is $40$.

(C) For a line to touch a circle,the perpendicular distance from the center of the circle to the line must be equal to the radius of the circle.
The equation of the circle is ${x^2} + {y^2} = {a^2}$,so its center is $(0, 0)$ and its radius is $a$.
The perpendicular distance from the center $(0, 0)$ to the line $x \cos \alpha + y \sin \alpha - p = 0$ is given by:
$d = \frac{|(0)\cos \alpha + (0)\sin \alpha - p|}{\sqrt{\cos^2 \alpha + \sin^2 \alpha}}$
Since $\cos^2 \alpha + \sin^2 \alpha = 1$,we have:
$d = \frac{|-p|}{\sqrt{1}} = |p|$
Setting the distance equal to the radius $(d = a)$:
$|p| = a$
Squaring both sides,we get:
${p^2} = {a^2}$

(B) The equation of the line is $3x - 2y = k$,which can be rewritten as $y = \frac{3}{2}x - \frac{k}{2}$.
Comparing this with $y = mx + c$,we get $m = \frac{3}{2}$ and $c = -\frac{k}{2}$.
The circle is ${x^2} + {y^2} = 4{r^2}$,which has a radius $a = 2r$.
$A$ line $y = mx + c$ is tangent to the circle ${x^2} + {y^2} = a^2$ if $c^2 = a^2(1 + m^2)$.
Substituting the values,we get $\left(-\frac{k}{2}\right)^2 = (2r)^2 \left(1 + \left(\frac{3}{2}\right)^2\right)$.
$\frac{k^2}{4} = 4{r^2} \left(1 + \frac{9}{4}\right)$.
$\frac{k^2}{4} = 4{r^2} \left(\frac{13}{4}\right)$.
$\frac{k^2}{4} = 13{r^2}$.
Therefore,${k^2} = 52{r^2}$.

(C) The equation of the tangent at $P(3, 4)$ to the circle ${x^2} + {y^2} = 25$ is given by $xx_1 + yy_1 = r^2$,which is $3x + 4y = 25$.
To find the intercepts on the coordinate axes,set $y = 0$ to get $3x = 25$,so $x = \frac{25}{3}$. Thus,the point $A$ is $(\frac{25}{3}, 0)$.
Set $x = 0$ to get $4y = 25$,so $y = \frac{25}{4}$. Thus,the point $B$ is $(0, \frac{25}{4})$.
The triangle formed by the tangent and the coordinate axes is a right-angled triangle $\Delta OAB$ with base $OA = \frac{25}{3}$ and height $OB = \frac{25}{4}$.
Area of $\Delta OAB = \frac{1}{2} \times OA \times OB = \frac{1}{2} \times \frac{25}{3} \times \frac{25}{4} = \frac{625}{24}$.

(C) The equation of the circle is $x^2 + y^2 = 16$,which is of the form $x^2 + y^2 = a^2$,where $a = 4$.
The line $y = mx + c$ is a tangent to the circle $x^2 + y^2 = a^2$ if $c^2 = a^2(1 + m^2)$.
Here,$m = 2$ and $a = 4$.
Substituting these values,we get $c^2 = 4^2(1 + 2^2) = 16(1 + 4) = 16(5) = 80$.
Therefore,$c = \pm \sqrt{80} = \pm 4\sqrt{5}$.
Since $4\sqrt{5}$ is one of the given options,the correct value is $4\sqrt{5}$.

(C) The given circle is $5x^2 + 5y^2 = 1$,which can be written as $x^2 + y^2 = \frac{1}{5}$.
Comparing this with $x^2 + y^2 = r^2$,we get $r^2 = \frac{1}{5}$,so $r = \frac{1}{\sqrt{5}}$.
The line parallel to $3x + 4y = 1$ is of the form $3x + 4y + k = 0$.
The distance from the center $(0, 0)$ to the tangent line is equal to the radius $r$.
Using the formula $d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$,we have $\frac{|k|}{\sqrt{3^2 + 4^2}} = \frac{1}{\sqrt{5}}$.
$\frac{|k|}{5} = \frac{1}{\sqrt{5}} \Rightarrow |k| = \frac{5}{\sqrt{5}} = \sqrt{5}$.
Thus,$k = \pm \sqrt{5}$.
The equations of the tangents are $3x + 4y = \pm \sqrt{5}$.

(A) For the circle $x^2 + y^2 = 1$,differentiating with respect to $x$ gives $2x + 2y \frac{dy}{dx} = 0$,which implies $\frac{dy}{dx} = -\frac{x}{y}$.
For a tangent to be parallel to the $x$-axis,the slope $\frac{dy}{dx}$ must be $0$.
Setting $\frac{dy}{dx} = 0$ gives $x = 0$.
Substituting $x = 0$ into the circle equation $x^2 + y^2 = 1$,we get $y^2 = 1$,so $y = \pm 1$.
Thus,there are two points $(0, 1)$ and $(0, -1)$ where the tangent is parallel to the $x$-axis.
Therefore,both $A$ and $R$ are true,and $R$ is the correct explanation of $A$.

(A) The given equation of the circle is $x^2 + y^2 - 2x + 6y - 6 = 0$.
Completing the square,we get $(x - 1)^2 + (y + 3)^2 = 6 + 1 + 9 = 16 = 4^2$.
Thus,the center of the circle is $(1, -3)$ and the radius $r = 4$.
The equation of the tangent parallel to $3x - 4y + 7 = 0$ is of the form $3x - 4y + k = 0$.
The perpendicular distance from the center $(1, -3)$ to the tangent line $3x - 4y + k = 0$ must be equal to the radius $r = 4$.
Using the distance formula $\frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} = r$,we have $\frac{|3(1) - 4(-3) + k|}{\sqrt{3^2 + (-4)^2}} = 4$.
$\frac{|3 + 12 + k|}{5} = 4$.
$|15 + k| = 20$.
This gives $15 + k = 20$ or $15 + k = -20$.
Therefore,$k = 5$ or $k = -35$.

(A) The equation of the circle is $(x - 1)^2 + (y - 2)^2 = r^2$,which has its center at $(1, 2)$ and radius $r$.
If a line $Ax + By + C = 0$ touches a circle,the perpendicular distance from the center of the circle to the line must be equal to the radius $r$.
The perpendicular distance $d$ from the point $(x_1, y_1)$ to the line $Ax + By + C = 0$ is given by $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.
Substituting the values,we get $r = \frac{|3(1) + 4(2) - 1|}{\sqrt{3^2 + 4^2}}$.
$r = \frac{|3 + 8 - 1|}{\sqrt{9 + 16}} = \frac{|10|}{\sqrt{25}} = \frac{10}{5} = 2$.
Thus,the value of $r$ is $2$.

(D) The given lines are ${x^2 - 3xy - 3x + 9y = 0}$,which can be factored as ${x(x - 3y) - 3(x - 3y) = 0}$,giving ${(x - 3)(x - 3y) = 0}$.
Thus,the normals are ${x = 3}$ and ${x = 3y}$.
The intersection of these lines gives the center of the required circle,which is ${(3, 1)}$.
The given circle is ${x^2 + y^2 - 6x + 6y + 17 = 0}$,which can be written as ${(x - 3)^2 + (y + 3)^2 = 1}$.
Its center is ${C_1 = (3, -3)}$ and radius ${r_1 = 1}$.
Let the required circle have center ${C_2 = (3, 1)}$ and radius ${r_2}$.
The distance between the centers is ${C_1C_2 = \sqrt{(3 - 3)^2 + (1 - (-3))^2} = \sqrt{0^2 + 4^2} = 4}$.
Since the circles touch externally,${C_1C_2 = r_1 + r_2}$.
${4 = 1 + r_2 \implies r_2 = 3}$.
The equation of the circle is ${(x - 3)^2 + (y - 1)^2 = 3^2}$.
${x^2 - 6x + 9 + y^2 - 2y + 1 = 9}$.
${x^2 + y^2 - 6x - 2y + 1 = 0}$.

(C) The equation of the tangent to the parabola $y = x^2 + 6$ at $(1, 7)$ is given by $\frac{1}{2}(y + 7) = x(1) + 6$.
Simplifying this,we get $y + 7 = 2x + 12$,which implies $y = 2x + 5$ $(i)$.
This tangent touches the circle $x^2 + y^2 + 16x + 12y + c = 0$ $(ii)$.
Substituting $(i)$ into $(ii)$:
$x^2 + (2x + 5)^2 + 16x + 12(2x + 5) + c = 0$
$x^2 + 4x^2 + 20x + 25 + 16x + 24x + 60 + c = 0$
$5x^2 + 60x + (85 + c) = 0$.
Since the line is tangent,the discriminant must be zero:
$D = (60)^2 - 4(5)(85 + c) = 0$
$3600 - 20(85 + c) = 0$
$180 - (85 + c) = 0 \implies c = 95$.
The quadratic equation becomes $5x^2 + 60x + 180 = 0$,or $x^2 + 12x + 36 = 0$.
$(x + 6)^2 = 0 \implies x = -6$.
Substituting $x = -6$ into $(i)$,$y = 2(-6) + 5 = -7$.
Thus,the point of contact is $(-6, -7)$.

(C) Given the curve equation is ${y^2} = 2(x - 3)$ .....$(i)$
Differentiating with respect to $x$,we get $2y \cdot \frac{dy}{dx} = 2$,which implies $\frac{dy}{dx} = \frac{1}{y}$.
The slope of the tangent at any point $(x, y)$ is $m_t = \frac{1}{y}$.
The slope of the normal is $m_n = -\frac{1}{m_t} = -y$.
The given line is $y - 2x + 1 = 0$,which can be written as $y = 2x - 1$. The slope of this line is $2$.
Since the normal is parallel to the line,their slopes must be equal:
$-y = 2 \implies y = -2$.
Substituting $y = -2$ into equation $(i)$:
$(-2)^2 = 2(x - 3)$
$4 = 2(x - 3)$
$2 = x - 3$
$x = 5$.
Therefore,the required point is $(5, -2)$.

(A) Given curve is $x^2 = 3 - 2y$ ...$(i)$
Differentiating both sides with respect to $x$,we get:
$2x = -2 \frac{dy}{dx}$
$\frac{dy}{dx} = -x$
The slope of the tangent to the curve at any point $(x, y)$ is $m = -x$.
The given line is $x + y = 2$,which can be written as $y = -x + 2$. The slope of this line is $-1$.
Since the line is tangent to the curve,the slope of the tangent must equal the slope of the line:
$-x = -1$
$x = 1$
Substituting $x = 1$ into the equation of the curve $(i)$:
$(1)^2 = 3 - 2y$
$1 = 3 - 2y$
$2y = 2$
$y = 1$
Therefore,the point of tangency is $(1, 1)$.

(C) The given equation of the circle is $x^2 + 2px + y^2 - 2qy + q^2 = 0$.
Rewriting this,we get $(x + p)^2 - p^2 + (y - q)^2 - q^2 + q^2 = 0$,which simplifies to $(x + p)^2 + (y - q)^2 = p^2$.
The center of the circle is $(-p, q)$ and the radius is $|p|$.
Since the circle passes through the origin $(0, 0)$ (as $0^2 + 2p(0) + 0^2 - 2q(0) + q^2 = q^2 \neq 0$,wait,let's re-evaluate).
Actually,the equation $x^2 + 2px + y^2 - 2qy + q^2 = 0$ at $(0,0)$ gives $q^2 = 0$,so $q=0$. If $q=0$,the circle is $x^2 + 2px + y^2 = 0$,which passes through the origin.
For the tangents from the origin to be perpendicular,the origin must lie on the circle,and the radius must be such that the angle between tangents is $90^\circ$.
Alternatively,the condition for perpendicular tangents from the origin to a circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is $g^2 + f^2 = 2c$.
Here,$2g = 2p \Rightarrow g = p$,$2f = -2q \Rightarrow f = -q$,and $c = q^2$.
Substituting these into the condition: $p^2 + (-q)^2 = 2(q^2)$ $\Rightarrow p^2 + q^2 = 2q^2$ $\Rightarrow p^2 = q^2$ $\Rightarrow p^2 - q^2 = 0$.

(C) The given circle is $x^2 + y^2 - 2x - 8 = 0$.
Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -1, f = 0, c = -8$.
The center is $(-g, -f) = (1, 0)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{(-1)^2 + 0^2 - (-8)} = \sqrt{1 + 8} = \sqrt{9} = 3$.
$A$ line $Ax + By + C = 0$ touches the circle if the perpendicular distance from the center to the line equals the radius.
The line is $3x + 4y - m = 0$.
Distance $d = \frac{|3(1) + 4(0) - m|}{\sqrt{3^2 + 4^2}} = 3$.
$\frac{|3 - m|}{\sqrt{9 + 16}} = 3$.
$|3 - m| = 3 \times 5 = 15$.
$3 - m = 15$ or $3 - m = -15$.
$m = 3 - 15 = -12$ or $m = 3 + 15 = 18$.
Thus,$m = 18, -12$.

(A) Let the circle be $S = x^2 + y^2 - 2x + 4y = 0$.
For the point $(x_1, y_1) = (0, 1)$,the power of the point is $S_1 = 0^2 + 1^2 - 2(0) + 4(1) = 5$.
The equation of the tangent $T$ at $(x_1, y_1)$ is given by $x x_1 + y y_1 - (x + x_1) + 2(y + y_1) = 0$.
Substituting $(0, 1)$,we get $T = x(0) + y(1) - (x + 0) + 2(y + 1) = -x + 3y + 2$.
The equation of the pair of tangents is given by $SS_1 = T^2$.
$5(x^2 + y^2 - 2x + 4y) = (-x + 3y + 2)^2$.
$5x^2 + 5y^2 - 10x + 20y = x^2 + 9y^2 + 4 - 6xy - 4x + 12y$.
$4x^2 - 4y^2 + 6xy - 6x + 8y - 4 = 0$.
Factoring this expression,we get $(2x - y + 1)(x + 2y - 2) = 0$.
Thus,the equations of the tangents are $2x - y + 1 = 0$ and $x + 2y - 2 = 0$.

(C) The given line is $lx + my + n = 0$.
The line is tangent to the circle $x^2 + y^2 = r^2$ if and only if the perpendicular distance from the center $(0, 0)$ to the line is equal to the radius $r$ of the circle.
The perpendicular distance $d$ from $(0, 0)$ to $lx + my + n = 0$ is given by:
$d = \frac{|l(0) + m(0) + n|}{\sqrt{l^2 + m^2}} = \frac{|n|}{\sqrt{l^2 + m^2}}$
Setting $d = r$:
$\frac{|n|}{\sqrt{l^2 + m^2}} = r$
Squaring both sides:
$\frac{n^2}{l^2 + m^2} = r^2$
Therefore,$n^2 = r^2(l^2 + m^2)$.

(C) The given equation of the circle is $x^2 + y^2 - 2x - 6y + 9 = 0$.
Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -1$ and $f = -3$.
The center of the circle is $(-g, -f) = (1, 3)$.
The radius of the circle is $r = \sqrt{g^2 + f^2 - c} = \sqrt{(-1)^2 + (-3)^2 - 9} = \sqrt{1 + 9 - 9} = \sqrt{1} = 1$.
Since the center is $(1, 3)$ and the radius is $1$,the circle touches the $y$-axis (which is the line $x = 0$) at the point where the $x$-coordinate is $0$ and the $y$-coordinate is the same as the center's $y$-coordinate.
Thus,the point of contact is $(0, 3)$.

(C) The circle is $(x - 7)^2 + (y + 1)^2 = 5^2$. The center is $C(7, -1)$ and the radius $r = 5$.
Let the distance from the origin $O(0, 0)$ to the center $C(7, -1)$ be $d$.
$d = \sqrt{7^2 + (-1)^2} = \sqrt{49 + 1} = \sqrt{50} = 5\sqrt{2}$.
Let $\theta$ be the angle between the tangents. Then $\sin(\theta / 2) = \frac{r}{d} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$.
Thus,$\theta / 2 = 45^\circ = \pi / 4$.
Therefore,$\theta = 2 \times (\pi / 4) = \pi / 2$.

(C) The line $y = x + c$ intersects the circle $x^2 + y^2 = 1$ at two coincident points if it 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) Let $C$ be the center of the circle and $T_1$ be the point of tangency. The center $C$ is $(-g, -f)$ and the radius $r = \sqrt{g^2 + f^2 - c}$.
In the right-angled triangle $\triangle PT_1C$,the angle at $P$ is $\frac{\theta}{2}$.
Thus,$\cot \frac{\theta}{2} = \frac{PT_1}{CT_1}$.
Here,$PT_1$ is the length of the tangent from $P(x_1, y_1)$ to the circle,which is $\sqrt{S_1} = \sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c}$.
$CT_1$ is the radius $r = \sqrt{g^2 + f^2 - c}$.
Therefore,$\cot \frac{\theta}{2} = \frac{\sqrt{S_1}}{\sqrt{g^2 + f^2 - c}}$.

(A) The equation of the circle is $x^2 + y^2 - 22x - 4y + 25 = 0$. The center is $(11, 2)$ and the radius $r = \sqrt{11^2 + 2^2 - 25} = \sqrt{121 + 4 - 25} = \sqrt{100} = 10$.
The line perpendicular to $5x + 12y + 8 = 0$ is of the form $12x - 5y + k = 0$.
Since the line is a tangent,the perpendicular distance from the center $(11, 2)$ to the line $12x - 5y + k = 0$ must be equal to the radius $r = 10$.
$\frac{|12(11) - 5(2) + k|}{\sqrt{12^2 + (-5)^2}} = 10$
$\frac{|132 - 10 + k|}{\sqrt{144 + 25}} = 10$
$\frac{|122 + k|}{13} = 10$
$|122 + k| = 130$
$122 + k = 130 \implies k = 8$
$122 + k = -130 \implies k = -252$
Thus,the equations of the tangents are $12x - 5y + 8 = 0$ and $12x - 5y - 252 = 0$,which is $12x - 5y = 252$.

(A) The equation of the circle is $x^2 + y^2 = a^2$.
Differentiating with respect to $x$,we get $2x + 2y \frac{dy}{dx} = 0$.
Thus,$\frac{dy}{dx} = -\frac{x}{y}$.
At the point $(h, k)$,the slope of the tangent is $m = -\frac{h}{k}$.

(A) The general equation of a tangent to the circle $x^2 + y^2 = a^2$ at any point $(a \cos \theta, a \sin \theta)$ is given by $x \cos \theta + y \sin \theta = a$.
For the circle $(x - 3)^2 + (y - 3)^2 = 1$,we can perform a translation of axes where $X = x - 3$ and $Y = y - 3$.
The equation becomes $X^2 + Y^2 = 1^2$,which is of the form $X^2 + Y^2 = a^2$ with $a = 1$.
The tangent to this circle is $X \cos \theta + Y \sin \theta = 1$.
Substituting back,we get $(x - 3) \cos \theta + (y - 3) \sin \theta = 1$.
Thus,both $(A)$ and $(R)$ are true,and $(R)$ provides the correct explanation for $(A)$.

(B) Let $PT$ and $PQ$ be the tangents drawn from point $P(\alpha, \beta)$ to the circle $x^{2} + y^{2} = a^{2}$,and let $\angle TPQ = \theta$.
If $O$ is the center of the circle,then $\angle TPO = \angle QPO = \theta/2$.
In the right-angled triangle $\triangle OTP$,we have:
$\tan(\theta/2) = \frac{OT}{PT} = \frac{a}{\sqrt{S_1}}$,where $S_1 = \alpha^{2} + \beta^{2} - a^{2}$.
Therefore,$\theta/2 = \tan^{-1}\left(\frac{a}{\sqrt{S_1}}\right)$.
Hence,$\theta = 2\tan^{-1}\left(\frac{a}{\sqrt{S_1}}\right)$.

(C) Let $P = (x_1, y_1)$. The equation of the tangent at $P$ is given by:
$xx_1 + yy_1 + 3(x + x_1) + 3(y + y_1) - 2 = 0$
$x(x_1 + 3) + y(y_1 + 3) + 3x_1 + 3y_1 - 2 = 0 \dots (i)$
Since $Q$ lies on the $y$-axis,its $x$-coordinate is $0$. Given $Q$ is the intersection of the tangent and the line $5x - 2y + 6 = 0$,we substitute $x = 0$ into the line equation:
$5(0) - 2y + 6 = 0 \implies 2y = 6 \implies y = 3$. So,$Q = (0, 3)$.
Since $Q$ lies on the tangent $(i)$,we substitute $(0, 3)$ into $(i)$:
$0(x_1 + 3) + 3(y_1 + 3) + 3x_1 + 3y_1 - 2 = 0$
$3y_1 + 9 + 3x_1 + 3y_1 - 2 = 0 \implies 3x_1 + 6y_1 + 7 = 0$.
The length $PQ^2 = (x_1 - 0)^2 + (y_1 - 3)^2 = x_1^2 + y_1^2 - 6y_1 + 9$.
Since $P(x_1, y_1)$ is on the circle,$x_1^2 + y_1^2 = 2 - 6x_1 - 6y_1$.
Substituting this into the expression for $PQ^2$:
$PQ^2 = (2 - 6x_1 - 6y_1) - 6y_1 + 9 = 11 - 6x_1 - 12y_1 = 11 - 2(3x_1 + 6y_1)$.
Using $3x_1 + 6y_1 = -7$:
$PQ^2 = 11 - 2(-7) = 11 + 14 = 25$.
Therefore,$PQ = 5$.

(B) The equation of the circle is $x^2 + y^2 - 5x + 2y - 48 = 0$.
Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get the center $C = (-g, -f) = (2.5, -1)$.
The normal to a circle at any point always passes through the center of the circle.
The slope of the normal passing through $(5, 6)$ and $(2.5, -1)$ is given by $m = \frac{6 - (-1)}{5 - 2.5} = \frac{7}{2.5} = \frac{14}{5}$.
The equation of the normal is $y - 6 = \frac{14}{5}(x - 5)$.
$5(y - 6) = 14(x - 5)$
$5y - 30 = 14x - 70$
$14x - 5y - 40 = 0$.

(B) The equation of the circle is $x^2 + y^2 - 2x - 4y - 20 = 0$. The center $A$ is $(1, 2)$ and the radius $r$ is $\sqrt{1^2 + 2^2 - (-20)} = \sqrt{1 + 4 + 20} = 5$.
The tangent at $B(1, 7)$ is given by $x(1) + y(7) - (x + 1) - 2(y + 7) - 20 = 0$,which simplifies to $x + 7y - x - 1 - 2y - 14 - 20 = 0$,or $5y = 35$,so $y = 7$.
The tangent at $D(4, -2)$ is given by $x(4) + y(-2) - (x + 4) - 2(y - 2) - 20 = 0$,which simplifies to $4x - 2y - x - 4 - 2y + 4 - 20 = 0$,or $3x - 4y = 20$.
Solving $y = 7$ and $3x - 4y = 20$ gives $3x - 4(7) = 20$,so $3x = 48$,which means $x = 16$. Thus,$C$ is $(16, 7)$.
The quadrilateral $ABCD$ consists of two congruent right-angled triangles $\triangle ABC$ and $\triangle ADC$. The area of $ABCD = 2 \times \text{Area}(\triangle ABC) = 2 \times (\frac{1}{2} \times AB \times BC) = AB \times BC$.
$AB = r = 5$.
$BC = \sqrt{(16 - 1)^2 + (7 - 7)^2} = \sqrt{15^2} = 15$.
Area of $ABCD = 5 \times 15 = 75$ square units.

(D) The given line is $12x - 5y + 9 = 0$. Its slope is $m_1 = \frac{12}{5}$.
Any line perpendicular to this must have a slope $m = -\frac{5}{12}$.
The circle is $x^2 + y^2 = 2^2$,so the radius $r = 2$.
The point of contact $(x_1, y_1)$ for a tangent with slope $m$ to the circle $x^2 + y^2 = r^2$ is given by $\left( \pm \frac{r^2 m}{\sqrt{r^2(1+m^2)}}, \mp \frac{r^2}{\sqrt{r^2(1+m^2)}} \right)$.
Alternatively,using the normal form,the point of contact is $\left( \pm \frac{r m}{\sqrt{1+m^2}}, \mp \frac{r}{\sqrt{1+m^2}} \right)$.
Here $m = -\frac{5}{12}$ and $r = 2$.
$\sqrt{1+m^2} = \sqrt{1 + \frac{25}{144}} = \sqrt{\frac{169}{144}} = \frac{13}{12}$.
$x_1 = \pm \frac{2 \times (-5/12)}{13/12} = \pm \frac{-10}{13} = \mp \frac{10}{13}$.
$y_1 = \mp \frac{2}{13/12} = \mp \frac{24}{13}$.
Wait,checking the calculation: The tangent equation is $y = mx \pm r\sqrt{1+m^2} \implies y = -\frac{5}{12}x \pm 2(\frac{13}{12}) \implies 12y = -5x \pm 26 \implies 5x + 12y = \pm 26$.
The point of contact for $ax + by + c = 0$ on $x^2 + y^2 = r^2$ is $(\frac{-ar^2}{c}, \frac{-br^2}{c})$.
For $5x + 12y - 26 = 0$,point is $(\frac{-5(4)}{-26}, \frac{-12(4)}{-26}) = (\frac{20}{26}, \frac{48}{26}) = (\frac{10}{13}, \frac{24}{13})$.
For $5x + 12y + 26 = 0$,point is $(-\frac{10}{13}, -\frac{24}{13})$.
Given the options,the intended answer is $D$.

(B) The given circle is $x^2 + y^2 = 25$,which has its center at the origin $(0, 0)$.
The perpendicular distance $d$ from the center $(0, 0)$ to the line $3x - 4y = 0$ is given by:
$d = \frac{|3(0) - 4(0)|}{\sqrt{3^2 + (-4)^2}} = \frac{0}{5} = 0$.
Since the perpendicular distance from the center to the line is $0$,the line passes through the center of the circle.
Any line passing through the center of a circle is a normal to that circle.
Therefore,the line $3x - 4y = 0$ is a normal to the circle $x^2 + y^2 = 25$.

(A) The center of the rectangle is the midpoint of the diagonal joining $(0, 0)$ and $(8, 6)$,which is $(\frac{0+8}{2}, \frac{0+6}{2}) = (4, 3)$.
The length of the diagonal is $\sqrt{(8-0)^2 + (6-0)^2} = \sqrt{64 + 36} = 10$.
The radius of the circumcircle is $R = \frac{10}{2} = 5$.
The equation of the circumcircle is $(x-4)^2 + (y-3)^2 = 5^2$,which simplifies to $x^2 + y^2 - 8x - 6y = 0$.
The slope of the diagonal is $m = \frac{6-0}{8-0} = \frac{6}{8} = \frac{3}{4}$.
The tangents parallel to the diagonal have the form $y = \frac{3}{4}x + c$,or $3x - 4y + 4c = 0$.
The distance from the center $(4, 3)$ to the tangent is equal to the radius $R = 5$.
Using the distance formula: $\frac{|3(4) - 4(3) + 4c|}{\sqrt{3^2 + (-4)^2}} = 5$.
$\frac{|12 - 12 + 4c|}{5} = 5 \implies |4c| = 25 \implies 4c = \pm 25$.
Substituting $4c$ back into the equation $3x - 4y + 4c = 0$,we get $3x - 4y \pm 25 = 0$.

(A) The given 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 $(-g, -f) = (3, -2)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{(-3)^2 + 2^2 - (-12)} = \sqrt{9 + 4 + 12} = \sqrt{25} = 5$.
The tangent is parallel to $4x + 3y + 5 = 0$,so its equation is of the form $4x + 3y + k = 0$.
The perpendicular distance from the center $(3, -2)$ to the tangent line $4x + 3y + k = 0$ must be equal to the radius $r = 5$.
Using the distance formula: $\frac{|4(3) + 3(-2) + k|}{\sqrt{4^2 + 3^2}} = 5$.
$\frac{|12 - 6 + k|}{\sqrt{16 + 9}} = 5$.
$\frac{|6 + k|}{5} = 5$.
$|6 + k| = 25$.
$6 + k = 25$ or $6 + k = -25$.
$k = 19$ or $k = -31$.
Thus,the equations of the tangents are $4x + 3y + 19 = 0$ and $4x + 3y - 31 = 0$.

(C) The given circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.
Its center is $C(-g, -f)$ and radius $r = \sqrt{g^2 + f^2 - c}$.
The given line is $(x + g) \cos \theta + (y + f) \sin \theta = k$,which can be rewritten as $x \cos \theta + y \sin \theta + (g \cos \theta + f \sin \theta - k) = 0$.
If the line is tangent to the circle,the perpendicular distance from the center $C(-g, -f)$ to the line must equal the radius $r$.
$r = \left| \frac{(-g) \cos \theta + (-f) \sin \theta + g \cos \theta + f \sin \theta - k}{\sqrt{\cos^2 \theta + \sin^2 \theta}} \right|$.
Simplifying the numerator,we get $r = \left| \frac{-k}{1} \right| = |k|$.
Thus,$\sqrt{g^2 + f^2 - c} = |k|$.
Squaring both sides,we get $g^2 + f^2 - c = k^2$,which simplifies to $g^2 + f^2 = k^2 + c$.

(B) The given circle equation is $x^2 + y^2 - 4x - 8y - 5 = 0$.
Comparing this with the standard form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -2$,$f = -4$,and $c = -5$.
The center of the circle is $(-g, -f) = (2, 4)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{(-2)^2 + (-4)^2 - (-5)} = \sqrt{4 + 16 + 5} = \sqrt{25} = 5$.
$A$ line $Ax + By + C = 0$ touches a circle if the perpendicular distance from the center to the line equals the radius.
The line is $3x - 4y - \lambda = 0$.
The distance $d = \frac{|3(2) - 4(4) - \lambda|}{\sqrt{3^2 + (-4)^2}} = \frac{|6 - 16 - \lambda|}{\sqrt{9 + 16}} = \frac{|-10 - \lambda|}{5}$.
Setting $d = r$,we get $\frac{|-10 - \lambda|}{5} = 5$,which implies $|-10 - \lambda| = 25$.
This gives two cases: $-10 - \lambda = 25 \Rightarrow \lambda = -35$ or $-10 - \lambda = -25 \Rightarrow \lambda = 15$.
Comparing with the given options,the correct value is $15$.