Question

The coordinates of a point $P$ on the line $x+y+13=0,$ nearest to the circle $x^2+y^2+4x+6y-5=0,$ are

• A. $\left(-15,2\right)$
• B. $\left(-5,-6\right)$
• C. $\left(-6,-7\right)$
• D. $\left(-7,-6\right)$

Answer 1

Answer: (C)

The distance of the centre $\left(-2,-3\right)$ from the line $x+y+13=0$ is $\left|\frac{-2-3+13}{\sqrt{2}}\right|=4\sqrt{2}$ units which is greater than the radius $3\sqrt{2}$
Therefore, the line is neither tangent nor secant to the circle
Hence, nearest point $P$ on the line from the circle is the foot of the perpendicular from the centre on the line.
$\Rightarrow$ If $P=\left(h,k\right)$ , then
$\frac{h+2}{1}=\frac{k+3}{1}=-\left(\frac{-2-3+13}{2}\right)$
$\Rightarrow h=-6$ &$k=-7$
$\Rightarrow P=\left(-6,-7\right)$