Question

If a circle whose centre is $(-1,1)$ touches the straight line $x+2y=12$,then the co-ordinates of the point of contact are

• A. $(-\frac{7}{2},-4)$
• B. $(\frac{6}{5},\frac{27}{5})$
• C. $(2,-7)$
• D. $(-2,-5)$

Answer 1

The correct option is B $(\frac{6}{5},\frac{27}{5})$

Let the center of circle be $O\equiv (-1,1)$ and the point of contact $p\equiv (a,b)$

Also, the given line ${}^{\prime }x+2y={12}^{\prime }$ passes through $P$ and touches the circle hence it is a tangent to the circle and the line $\overrightarrow{OP}$ is normal to it.

$a+2b=12------\left(i\right)$

$x+2y=12\Rightarrow y=\frac{-x}{2}+6-------\left(ii\right)$

Here slope of line $=\frac{-1}{2}$

Since slope of line $\perp$ to a given line $=\frac{-1}{\left(Slopeofgivenline\right)}$

$\therefore$ Slope of normal $=\frac{-1}{(-1/2)}=2$.

$\therefore$ Slope of $\overrightarrow{OP}=2$

$\therefore \frac{b-1}{a-(-1)}=2$

$b-1=2a+2\Rightarrow b=2a+3$

From $\left(i\right)$

$a+2(2a+3)=12$

$5a+6=12$

$a=\frac{6}{5}$

$b=2\left(\frac{6}{5}\right)+3=\frac{12}{5}+3=\frac{27}{5}$