Question

State the center coordinate and radius length of the circle: x^2 + y^2 - 8x + 12y + 40 = 0

Answer 1

The circle is centered at (4,-6) and has a radius of √12.

Explanation:

The standard form for the circle equation is:

`(x-a)^2+(y-b)^2=r^2`

You have:

`x^2+y^2-8x+12y+40=0`

First complete the square for the 'x' terms:

`(x^2-8x+40)+y^2+12y=0`

`(x^2-8x+16-16+40)+y^2+12y=0`

`(x^2-8x+16)+y^2+12y+24=0`

`(x-4)^2+y^2+12y+24=0`

The 'x' terms are now in standard form. Work on the 'y' terms by completing the square once again:

`(x-4)^2+(y^2+12y+36-36+24)=0`

`(x-4)^2+(y^2+12y+36)-12=0`

`(x-4)^2+(y+6)^2-12=0`

Add the 12 to the other side:

`(x-4)^2+(y+6)^2=12`

The circle is centered at x = 4, y = -6, and has a radius of √12.

Answer 2

Below in bold.

Explanation:

x^2 + y^2 - 8x + 12y + 40 = 0

x^2 - 8x + y^2 + 12y = -40

Complete the square on the x and y terms:-

(x - 4)^2 - 16 + ( y + 6)^2 - 36 = -40

(x - 4)^2 + (y + 6)^2 = -40 + 16 + 36

(x - 4)^2 + (y + 6)^2 = 12

So the centre is at (4, -6) and the radius is √12.