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Q.
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
NTA AbhyasNTA Abhyas 2022
Solution:
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
$\frac{h + 2}{1}=\frac{k + 3}{1}=-\left(\frac{- 2 - 3 + 13}{2}\right)$
$\Rightarrow h=-6$ &$k=-7$
$\Rightarrow P=\left(- 6 , - 7\right)$