Question

The equation of a circle is x2+y2+8x−14y+56=0 . What is the radius of the circle?

Answer 1

Answer: $r=3$

Step-by-step explanation:

The equation of the circle in center-radius form is:

$(x-h)^2+(y-k)^2=r^2$

Where the point $(h,k)$ is the center of the circle and "r" is the radius.

Subtract 56 from both sides of the equation:

$x^2+y^2+8x-14y+56-56=0-56\\x^2+y^2+8x-14y=-56$

Make two groups for variable "x" and variable "y":

$(x^2+8x)+(y^2-14y)=-56$

Complete the square:

Add $(\frac{8}{2})^2=4^2$ inside the parentheses of "x".

Add  $(\frac{14}{2})^2=7^2$ inside the parentheses of "y".

Add $4^2$ and $7^2$ to the right side of the equation.

Then:

$(x^2+8x+4^2)+(y^2-14y+7^2)=-56+4^2+7^2\\\\(x^2+8x+4^2)+(y^2-14y+7^2)=9$

Rewriting, you get that the equation of the circle in center-radius form is:

$(x+4)^2+(y-7)^2=3^2$

You can observe that the radius of the circle is:

$r=3$