Question

The mid point of the chord of the circle $x^2+y^2-6x+4y-12=0$ drawn parallel to the tangent at $(-1,1)$ and at a distance of one unit from the tangent is

• A. $\left(\frac{3}{4},\frac{1}{4}\right)$
• B. $\left(\frac{1}{4},\frac{3}{4}\right)$
• C. $\left(-\frac{1}{5},\frac{2}{5}\right)$
• D. $\left(\frac{3}{5},\frac{2}{5}\right)$

Answer 1

Answer: (C)

Equation of circle is
$x^2+y^2-6x+4y-12=0$
$\therefore 2x+2y{y}^{\prime }-6+4y^{\prime }\&=0$
$\Rightarrow y^{\prime }=\frac{6-2x}{2y+4}$
$\therefore$ Slope of tangent at $(-1,1)={\frac{dv}{dx}|}_{(-1,1)}$ $=\frac{6-2(-1)}{2(1)+4}=\frac{8}{6}=\frac{4}{3}$
$\therefore$ Equation of tangent is
$y-1=\frac{4}{3}(x+1)$
$\Rightarrow 3y-3=4x+4$
$\Rightarrow 4x-3y+7=0$
Let the equation of chord is
$4x-3y+k=0$
$\therefore \frac{k-7}{\sqrt{16+9}}=\pm 1$
$\Rightarrow k-7=\pm 5$
$\Rightarrow k=7\pm 5\Rightarrow k=12,2$
$\therefore$ Equation of chord can be
$4x-3y+12=0$ or $4x-3y+2=0$
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 $4x-3y+2=0$
Now, slope of line $4x-3y+2=0$ is $\frac{4}{3}$
$\therefore$ Slope of line $CP=\frac{-3}{4}$
$\therefore$ Equation of $CP$ is
$y+2=\frac{-3}{4}(x-3)$
$\Rightarrow 4y+8=-3x+9$
$\Rightarrow 3x+4y=1$
On solving $4x-3y+2=0$ and $3x+4y=1$,
we get $P\left(-\frac{1}{5},\frac{2}{5}\right)$
$\therefore$ Mid-point of chord is $\left(-\frac{1}{5},\frac{2}{5}\right)$