The sum of the minimum and maximum distance of the point $(4,-3)$ to the circle $x^2+y^2+4 x-10 y-7=0$, is
The sum of the minimum and maximum distance of the point $(4,-3)$ to the circle $x^2+y^2+4 x-10 y-7=0$, is
$10$
$12$
$16$
$20$
Solution
Let the given point be $P(4,-3)$ and the given circle is
$x^2+y^2+4 x-10 y-7=0$
Centre of circle $=\mathrm{C}(-2,5)$
$\text { Radius }=\sqrt{(-2)^2+(5)^2+7}$
$=\sqrt{4+25+7}=\sqrt{36}=6$
Maximum distance, $a=C P+r$
Minimum distance, $b=C P-r$
Sum of the maximum and minimum distance,
$a+b=C P+r+C P-r=2 C P$
and $C P=\sqrt{(-2-4)^2+(5+3)^2}$
$=\sqrt{36+64}=\sqrt{100}=10$
Thus, $\quad a+b=2 C P=2 \times 10=20$