Question

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

• A. 13
• B. $2\sqrt{5}$
• C. 12
• D. 5

Answer 1

Answer: (B)

Given circle, $x^2+y^2+6x+4y-12=0$
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 $AP=d$
$AP=OP-OA$
$(OA=R)$
$OP=\sqrt{(-3-2^2+(-2+14{)}^2}$
$=\sqrt{25+144}=\sqrt{169}=13$
$\Rightarrow AP=13-5=8=d$
Length of tangent from the point $P(2,-14)$ is
$=\sqrt{(2{)}^2+(-14{)}^2+6\cdot 2+4(-14)-12}$
$=\sqrt{4+196+12-56-12}$
$=\sqrt{144}=12=l$
So, required $\sqrt{d+l}=\sqrt{8+12}=\sqrt{20}=2\sqrt{5}$.