# Need to find Center and Radius of this equation: x^2+y^2+10y-56=0Listen, I know how to do these, with the completing the square and everything but, I'm so confused because there is two y's but only...

Need to find Center and Radius of this equation: x^2+y^2+10y-56=0

Listen, I know how to do these, with the completing the square and everything but, I'm so confused because there is two y's but only one x, how am i supposed to complete the square on the x if i only have x^2?!?!?!??!

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### 2 Answers

The center and radius of the circle x^2 + y^2 + 10y - 56 = 0 has to be determined.

x^2 + y^2 + 10y - 56 = 0

=> x^2 + y^2 + 10y + 25 - 25 - 56 = 0

=> x^2 + y^2 + 10y + 25 = 81

=> x^2 + (y + 5)^2 = 9^2

This is in the form (x - h)^2 + (y - k)^2 = r^2 of a circle where the center is (h, k) and the radius is r

**The center of the circle is (0, -5) and the radius is 9.**

The general equarion for a circle with center (x1,y1) and rarius r is:

(x-x1)^2 + (y-y1)^2 = r^2

we bring the given equation x^2+y^2+10y-56=0 to the above form as:

x^2 + y^2 + 10y = 56

x^2 + y^2 +10y +25 = 56 + 25

x^2 + (y^2 +10y +25) = 81

(x-0)^2 + (y+5)^2 = (9)^2

this gives:

x1 = 0 and y1 = -5

*The center of the circle with given equation is (0,-5)*