Question

State whether the given statement is True or False. The shortest distance from the point $(2,-7)$ to the circle $x^2+y^2-14x-10y-151=0$ is equal to $5$.

Answer 1

Given circle:
$x^2+y^2-14x-10y-151=0\dots \left(1\right)$
General equation of circle:
$x^2+y^2+2gx+2fy+c=0\dots \left(2\right)$
Comparing the equations $\left(1\right)$ and $\left(2\right)$ we get
$g=-7,f=-5,c=-151$
$\Rightarrow$ Centre of given circle $C(7,5)$
[$\because$ Centre of circle is $(-g,-f)$ ] And radius is
$r=\sqrt{g^2+f^2-c}=\sqrt{49+25+151}$
$\Rightarrow r=15$
Distance from point $(2,-7)$ to centre of circle $(7,5)$ is $\sqrt{(7-2{)}^2+(5+7{)}^2}=13$
Since this distance is less than the radius of the circle, the point is inside the circle
Hence the shortest distance of the point $(2,-7)$ from the given circle is the difference between radius and distance of point from centre of the circle.

Hence shortest distance is $=15-13=2$
Hence, the Statement is false