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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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
x1 = 0 and y1 = -5
The center of the circle with given equation is (0,-5)
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