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Q.
The mid point of the chord of the circle $x^{2}+y^{2}-6 x+4 y-12=0$ drawn parallel to the tangent at $(-1,1)$ and at a distance of one unit from the tangent is
TS EAMCET 2020
Solution:
Equation of circle is
$x^{2}+y^{2}-6 x+4 y-12=0 $
$\therefore 2 x+2 y y^{\prime}-6+4 y^{\prime} \&=0$
$\Rightarrow y^{\prime}=\frac{6-2 x}{2 y+4}$
$\therefore $ Slope of tangent at $(-1,1)=\left.\frac{d v}{d x}\right|_{(-1,1)}$ $=\frac{6-2(-1)}{2(1)+4}=\frac{8}{6}=\frac{4}{3}$
$\therefore $ Equation of tangent is
$y-1=\frac{4}{3}(x+1) $
$\Rightarrow 3 y-3=4 x+4 $
$\Rightarrow 4 x-3 y+7=0$
Let the equation of chord is
$4 x-3 y+k=0$
$\therefore \frac{k-7}{\sqrt{16+9}}=\pm 1 $
$\Rightarrow k-7=\pm 5$
$\Rightarrow k=7 \pm 5 \Rightarrow k=12,2$
$\therefore $ Equation of chord can be
$4 x-3 y+12=0$ or $4 x-3 y+2=0$
Now, Distance of both above lines from centre of circle are $d_{1}=\left|\frac{12+6+12}{5}\right|=6$
and $d_{2}=\left|\frac{12+6+2}{5}\right|=4$
and radius of circle, $r=\sqrt{9+4+12}=5$
$ \therefore d_{2} $ Equation of chord is $4 x-3 y+2=0$
Now, slope of line $4 x-3 y+2=0$ is $\frac{4}{3}$
$\therefore $ Slope of line $C P=\frac{-3}{4}$
$\therefore $ Equation of $C P$ is
$y+2=\frac{-3}{4}(x-3) $
$\Rightarrow 4 y+8=-3 x+9$
$ \Rightarrow 3 x+4 y=1$
On solving $4 x-3 y+2=0$ and $3 x+4 y=1$,
we get $P\left(-\frac{1}{5}, \frac{2}{5}\right)$
$\therefore $ Mid-point of chord is $\left(-\frac{1}{5}, \frac{2}{5}\right)$