Equation of the given circle is $x^2+y^2-14x-10y-151=0$
The given point is $P(2,-7)$
The shortest distance =5
The coordinates of the centre is $(7,5)$
$\therefore$ Radius of the circle is $\sqrt{49+25+151}$
$\Rightarrow \sqrt{225}=15$
Distance between the centre $'O'$ and the point $P(2,-7)$ is
$OP=\sqrt{(7-2)^2+(5+7)^2}$
$\;\;\;\;\;=\sqrt{25+144}$
$\;\;\;\;\;=\sqrt{169}$
$\;\;\;\;\;=13$
Hence the shortest distance $=15-13=2$
Hence the statement is False.