Maths Solve the Following Question.(2 Marks)

Comparing this equation with $x^2+y^2=a^2$, we get

$a^2=9$

The locus of the point of intersection of perpendicular tangents is the director circle of the given circle.

The equation of the director circle of the circle $x^2+y^2=a^2$ is $x^2+y^2=2 a^2$.

the required equation is

$\begin{aligned} & x^2+y^2=2(9) \\ & x^2+y^2=18\end{aligned}$

Alternate method:

Given equation of the circle is $x^2+y^2=9$

Comparing this equation with $x^2+y^2=a^2$, we get $a^2=9$

Let $P\left(x_1, y_1\right)$ be a point on the required locus.

Equations of the tangents to the circle $x^2+y^2=a^2$ with slope $m$ are

$y=m x \pm \sqrt{a^2\left(1+m^2\right)}$

∴ Equations of the tangents are

$\begin{aligned} & y=m x \pm \sqrt{9}\left(m^2+1\right) \\ & \Rightarrow y=m x \pm 3 \sqrt{1+m^2}\end{aligned}$

Since, these tangents pass through (x1, y1).

$\begin{aligned} & \mathrm{y}_1=\mathrm{mx_{1 }} \pm 3 \sqrt{1+m^2} \\ & \Rightarrow \mathrm{y}_1-\mathrm{mx_{1 }}= \pm 3 \sqrt{1+m^2} \\ & \Rightarrow\left(\mathrm{y}_1-\mathrm{mx_{1 }}\right)^2=9\left(1+\mathrm{m}^2\right) \ldots \ldots[\text { Squaring both the sides] } \\ & \Rightarrow y_1^2-2 m x_1 y_1+m^2 x_1^2=9+9 m^2 \\ & \Rightarrow\left(x_1^2-9\right) \mathrm{m}^2-2 m x_1 y_1+\left(y_1^2-9\right)=0\end{aligned}$

This is a quadratic equation which has two roots m1 and m2.

$m_1 m_2=\frac{y_1^2-9}{x_1^2-9}$

Since, the tangents are at right angles. m1m2 = -1

$\begin{aligned} & \Rightarrow \frac{y_1^2-9}{x_1^2-9}=-1 \\ & \Rightarrow y_1^2-9=9-x_1^2 \\ & \Rightarrow x_1^2+y_1^2=18\end{aligned}$

Equation of the locus of point $P$ is $x^2+y^2=18$.

Comparing this equation with $x^2+y^2=a^2$, we get

$a^2=4$

Given equation of the line is $3 x+2 y+1=0$

Slope of this line $=\frac{-3}{2}$

Since, the required tangents are parallel to the given line.

Slope of required tangents $(\mathrm{m})=\frac{-3}{2}$

Equations of the tangents to the circle $x^2+y^2=a^2$ with slope $m$ are

$y=m x \pm \sqrt{a^2\left(1+m^2\right)}$

the required equations of the tangents are

$\begin{aligned} y & =\frac{-3}{2} x \pm \sqrt{4\left[1+\left(\frac{-3}{2}\right)^2\right]} \\ & =\frac{-3}{2} x \pm \sqrt{4\left(1+\frac{9}{4}\right)}\end{aligned}$

$y=\frac{-3}{2} x \pm \sqrt{13}$

$\begin{aligned} & 2 y=-3 x \pm 2 \sqrt{13} \\ & 3 x+2 y \pm 2 \sqrt{13}=0\end{aligned}$

Comparing this equation with $x^2+y^2=r^2$, we get

$r^2=5$

The equation of a tangent to the circle $x^2+y^2=r^2$ at $\left(x_1, y_1\right)$ is $x x_1+y y_1=r^2$

the equation of the tangent at (1, -2) is

x(1) + y(-2) = 5

⇒ x – 2y = 5

Comparing this equation with $x^2+y^2+2 g x+2 f y+c=0$, we get

2g = -6, 2f = -4, c = 0

⇒ g = -3, f = -2, c = 0

The equation of a tangent to the circle $x^2+y^2+2 g x+2 f y+c=0$ at $\left(x_1, y_1\right)$ is

$x x_1+y y_1+g\left(x+x_1\right)+f\left(y+y_1\right)+c=0$

the equation of the tangent at (6, 4) is

$\begin{aligned} & x(6)+y(4)-3(x+6)-2(y+4)+0=0 \\ & \Rightarrow 6 x+4 y-3 x-18-2 y-8=0 \\ & \Rightarrow 3 x+2 y-26=0\end{aligned}$

Alternate method:

Given equation of the circle is $x^2+y^2-6 x-4 y=0$

x(x – 6) + y(y – 4) = 0, which is in diameter form where (0, 0) and (6, 4) are endpoints of diameter.

Slope of $O P=\frac{4-0}{6-0}=\frac{2}{3}$

Since, OP is perpendicular to the required tangent.

Slope of the required tangent $=\frac{-3}{2}$

the equation of the tangent at (6, 4) is

$\begin{aligned} & y-4=\frac{-3}{2}(x-6) \\ & \Rightarrow 2(y-4)=3(x-6) \\ & \Rightarrow 2 y-8=-3 x+18 \\ & \Rightarrow 3 x+2 y-26=0\end{aligned}$

Here, $g=-1, f=\frac{\mathbf{k}}{2}, c=-23$

Length of the tangent segment to the circle $x^2+y^2+2 g x+2 f y+c=0$ from the point $\left(x_1\right.$,$\left.y_1\right)$ is $\sqrt{x_1^2+y_1^2+2 g x_1+2 f y_1+c}$

Length of the tangent segment from (8, -3) = √10

$\begin{aligned} & \Rightarrow \sqrt{8^2+(-3)^2-2(8)+k(-3)-23}=\sqrt{10} \\ & \Rightarrow 64+9-16-3 k-23=10 \ldots .[\text { Squaring both the sides] } \\ & \Rightarrow 34-3 k=10 \\ & \Rightarrow 3 k=24 \\ & \Rightarrow k=8\end{aligned}$

Comparina this equation with $x^2+y^2=r^2$ we aet $r=8$

The equation of a tangent to the circle $x^2+y^2=r^2$ at $P(\theta)$ is $x \cos \theta+y \sin \theta=r$

$\therefore$ the equation of the tangent at $P\left(\frac{2 \pi}{3}\right)$ is

$\begin{aligned} & \Rightarrow x \cos \frac{2 \pi}{3}+y \sin \frac{2 \pi}{3}=9 \\ & \Rightarrow x\left(\frac{-1}{2}\right)+y\left(\frac{\sqrt{3}}{2}\right)=8 \\ & \Rightarrow-x+\sqrt{3} y=16 \\ & \Rightarrow x-\sqrt{3} y+16=0\end{aligned}$

⇒ x – 3 = -4 sin θ, y – 2 = -4 cos θ

On squaring and adding, we get

$\begin{aligned} & \Rightarrow(x-3)^2+(y-2)^2=(-4 \sin \theta)^2+(-4 \cos \theta)^2 \\ & \Rightarrow(x-3)^2+(y-2)^2=16 \sin ^2 \theta+16 \cos ^2 \theta \\ & \Rightarrow(x-3)^2+(y-2)^2=16\left(\sin ^2 \theta+\cos ^2 \theta\right) \\ & \Rightarrow(x-3)^2+(y-2)^2=16(1) \\ & \Rightarrow(x-3)^2+(y-2)^2=16 \\ & \Rightarrow(x-3)^2+(y-2)^2=4^2\end{aligned}$

Comparing this equation with $(x-h)^2+(y-k)^2=r^2$, we get

h = 3, k = 2, r = 4

∴ Centre of the circle is (3, 2) and radius is 4.

Comparing this equation with $x^2+y^2+2 g x+2 f y+c=0$, we get

2g = -1, 2f = 2 and c = -3

$g=\frac{-1}{2}, f=1$ and $c=-3$

Centre of the circle $=(-g,-f)=\left(\frac{1}{2},-1\right)$

and radius of the circle

$\begin{aligned} & =\sqrt{g^2+\mathrm{f}^2-\mathrm{c}} \\ & =\sqrt{\left(-\frac{1}{2}\right)^2+(1)^2-(-3)}\end{aligned}$

$\begin{aligned} & =\sqrt{\frac{17}{4}} \\ & =\frac{\sqrt{17}}{2}\end{aligned}$

centre of the circle $\mathrm{C}(h, k)=\mathrm{C}(3,-4)$ $3 x-4 y-5=0$ cuts the circle at $\mathrm{A}$ and $\mathrm{B}$.
$
\begin{aligned}
& l(\mathrm{AB})=6 \\
& \mathrm{CM} \perp \mathrm{AB}
\end{aligned}
$
$\therefore \mathrm{AM}=\mathrm{BM}=3$
$\mathrm{CM}=$ Length of perpendicular from centre on the line
$
\begin{aligned}
& =\left|\frac{3(3)-4(-4)-5}{\sqrt{(3)^2+(-4)^2}}\right| \\
& =\left|\frac{9+16-5}{\sqrt{9+16}}\right| \\
& =\left|\frac{20}{5}\right|=4
\end{aligned}
$

The circle touches the $\mathrm{Y}$-axis at point $(0,3)$, and the centre is $(-3,3)$ we get radius $r=3$
By using centre radius form;
$
\begin{aligned}
& (x-h)^2+(y-k)^2=r^2 \\
& (x+3)^2+(y-3)^2=9 \\
& x^2+6 x+9+y^2-6 y+9=9
\end{aligned}
$
$\therefore x^2+y^2+6 x-6 y+9=0$ is the equation of the circle.

$x^2+y^2=r^2$

Here, r = 4

$\therefore$ The required equation of the circle is $\mathrm{x}^2+\mathrm{y}^2=4^2$ i.e., $\mathrm{x}^2+\mathrm{y}^2=16$.