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Q.
If a circle, whose centre is $(-1,1)$ touches the straight line $x+2 y$ $+12=0$, then the coordinates of the point of contact are
Conic Sections
Solution:
Let point of contact be $P\left(x_{1}, y_{1}\right)$.
This point lies on line
$x_{1}+2 y_{1}=-12$ ... (i)
Gradient of $P O=m_{1}=\frac{y_{1}-1}{x_{1}+1}$
Gradient of $x+2 y+12=m_{2}=-\frac{1}{2}$
The two lines are perpendicular
$\therefore m_{1} m_{2}=-1$
$\Rightarrow \left(\frac{y_{1}-1}{x_{1}+1}\right)\left(\frac{-1}{2}\right)=-1$
$ \Rightarrow y_{1}-1=2 x_{1}+2$
$\Rightarrow 2 x_{1}-y_{1}=-3$ ... (ii)
On solving equation (i) and (ii), we get
$\left(x_{1}, y_{1}\right)=\left(\frac{-18}{5}, \frac{-21}{5}\right)$
