Question

Compute the shortest distance between the circle $x^2+y^2-10x-14y-151=0$ and the point $(-7,2)$.

• A. $0$
• B. $1$
• C. $2$
• D. $4$

Answer 1

The correct option is C $2$

$x^2+y^2-10x-14y-151=0$
Putting $(-7,2)$ in circle
$49+4+70-28-151<0$
So, point lies inside the circle
${(x-5)}^2+{(y-7)}^2=225$
$r=15$, $c=(5,7)$
SD = Radius - distance between $(5,7)$ and $(-7,2)$
$=15-\sqrt{144+25}$
$=15-13=2$ units