How do you find the center and radious of this equation?

In general form, the equation of a circle on its plane is

x^2 + y^2 + ax + by + c = 0

This can be written as

(x + a/2)^2 + (y +b/2)^2 + c - (a^2)/4 - (b^2)/4 = 0

(x + a/2)^2 + (y + b/2)^2 = (a^2 + b^2 - 4c)/4

So, the center is at (-a/2, -b/2) and the radius is r = (sqrt(a^2 + b^2 - 4c))/2, provided a^2 + b^2 - 4c > 0. If it is 0, the circle reduces to its center.

In your case, the center is at (5, 6) and the radius is sqrt(100 + 144 + 32)/2 = sqrt(276)/2 = sqrt(39)

We have to complete the square... x^2 - 10x (add and subtract 25) y^2 -12y ( add and subtract 36)

It is x^2 - 10x + 25 -25 y^2 -12y +36 -36 - 8 = 0 -->

(x -5)^2 + (y -6)^2 = 69 ...

In the standard form the center is (h,k) and the radius is sqrt(r^2)

then, the center is (5,6) and the radius is r = sqrt(69) OK!

You have do something called completing the square for both the "x"s and the "y"s:

x² - 10x + 25 + y² - 12y + 36 - 8 = 25 + 36

(x - 5)² + (y - 6)² -8 = 61

(x - 5)² + (y - 6)² = 69

Thus the coordinates of the centre are (5, 6) and the radius is √69.

x^2 + y^2 - 10x - 12y - 8 = 0

x^2 + y^2 - 10x - 12y = 8

Complete the square of x and y by adding 25 and 36 as shown below

(x^2 - 10x +25)+(y^2- 12y+36) = 8+25+36

(x-5)^2 +(y-6)^2 =69 = {sqrt(69)}^2

Here h =5 and k =6 and r = sqrt(69)

Hence centre are(5,6) and radius = sqrt(69).......................Ans

If the standard equation of the circle is x^2+y^2+2gx+2fy+c=0, then centre = (-g,-f) and radius = sqrt(g^2+f^2-c). Thus for your problem Centre = ( 5, 6) and radius = sqrt(69).