The least distance of the point $(10,7)$ from the circle $x^2+y^2-4 x-2 y-20=0$ is
The least distance of the point $(10,7)$ from the circle $x^2+y^2-4 x-2 y-20=0$ is
$6$
$7$
$4$
$5$
Solution
Given equation $\left(x^2+y^2-4 x-2 y-20=0\right)$
So, $C=(2,1)$ and radius $=\sqrt{(g)^2+(f)^2-c}$
$=\sqrt{(2)^2+(1)^2+20}$
Radius $=5$
When substituted $x=10$ and $y=7$ in equation,
then value becomes
$(10)^2+(7)^2-4(10)-2(7)-20=75$
which is greater than zero.
Thus, the point $(10,7)$ lies outside the circle.
Its distance from the centre of the circle $(2,1)$ is
$\sqrt{(10-2)^2+(7-1)^2}=10$ units
So, the minimum distance from the circle therefore becomes $10-5=5$ units