If the shortest distance from $(2,-14)$ to the circle $x^2+y^2+6 x+4 y-12=0$ is $d$ and the length of the tangent drawn from the same point to the circle is $l$, then $\sqrt{d+l}=$

Question

Aicharya · Accepted Answer

Given circle, $x^2+y^2+6 x+4 y-12=0$ Here, centre is $O(-3,-2)$ and radius $=\sqrt{9+4+12}=\sqrt{25}=5$

Shortest distance of the point $P(2,-14)$ to the circle is $ \text { circle is } \begin{aligned} A P & =d \\ A P & =O P-O A \\ (O A & =R) \\ O P & =\sqrt{(-3-2)^2+(-2+14)^2} \\ & =\sqrt{25+144}=\sqrt{169}=13 \\ \Rightarrow \quad A P & =13-5=8=d \end{aligned} $ Length of tangent from the point $P(2,-14)$ is $ \begin{aligned} & =\sqrt{(2)^2+(-14)^2+6 \cdot 2+4(-14)-12} \\ & =\sqrt{4+196+12-56-12} \\ & =\sqrt{144}=12=1 \end{aligned} $ So, required $\sqrt{d+l}=\sqrt{8+12}=\sqrt{20}=2 \sqrt{5}$

If the shortest distance from $(2,-14)$ to the circle $x^2+y^2+6 x+4 y-12=0$ is $d$ and the length of the…

Solution