Question

The sum of the minimum distance and the maximum distance from the point $(4,-3)$ to the circle
$x^2+y^2+4x-10y-7=0$ is

• A. $20$
• B. $12$
• C. $10$
• D. $16$

Answer 1

Answer: (A)

Centre of the given circle $\equiv C\left(-2,5\right)$
Radius of the circle $CN=CT=\sqrt{g^2+f^2-c}$
$=\sqrt{2^2+5^2+7}=\sqrt{36}=6$
Distance between $\left(4,-3\right)$ and $\left(-2,5\right)$is
$PC=\sqrt{6^2+8^2}=\sqrt{100}=10$
We join the external point,$\left(4,-3\right)$ to the centre of the circle $\left(-2,5\right)$.

Then $PT$ is the minimum distance, from external point $P$ to the circle and $PN$ is the maximum distance.
Minimum distance $=PT=PC-CT=10-6=4$
Maximum distance $=PN=PC+CN=10+6=16$
So, sum of minimum and maximum distance $=16+4=20$.