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`

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
`((8)/(2))^2=4^2` inside the parentheses of "x".

Add
`((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`