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 $= 16 + 4 = 20$.