Question

Find the shortest distance from the points $M(-7,2)$ to the circle $x^2+y^2-10x-14y-151=0$

Answer 1

Let the equation of the circle be,

$S=x^2+y^2-10x+14y-151=0$

Substituting the point $\left(-7,2\right)$ in the given equation,

$S={\left(-7\right)}^2+{\left(2\right)}^2-10\left(-7\right)+14\left(2\right)-151$

$=0$

Therefore, $M\left(-7,2\right)$ is on the circle.

The general equation of the circle is,

$a{x}^2+b{y}^2+2gx+2fy+c=0$

Comparing the given equation by general equation of the circle,

$g=-5$, $f=7$ and $c=-151$.

The radius of the circle is,

$r=\sqrt{g^2+f^2-c}$

$=\sqrt{{\left(-5\right)}^2+7^2+151}$

$=15$

The center of the circle is,

$C=\left(-g,-f\right)$

$=\left(5,-7\right)$

Now,

$CM=\sqrt{{\left(5+7\right)}^2+{\left(-7-2\right)}^2}$

$=\sqrt{144+81}$

$=\sqrt{225}$

$=15$

The shortest distance is,

$r-CM=15-15$

$=0$