Find the center and radius

## Find the center and the radius of the circle: 2x^2+2y^2-14x+4y-14=0

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# 1 Answer

First, note that (x + a)^2 = x^2 + 2ax + a^2.

We will complete the square to get x^2 + 2ax + a^2 and replace it with (x + a)^2.

2x^2+2y^2-14x+4y-14=0

Divide both sides by 2:

x^2+y^2-7x+2y-7=0

Rearrange terms toward getting the patterns we want:

x^2-7x+y^2+2y=7

Force the perfect square in x and in y:

x^2 + 2(-7/2)x + (-7/2)^2 - (-7/2)^2 + y^2+2(1)y + (1)^2 - (1)^2 = 7

The x term was restructured as 2ax to identify a. Then a^2 was added to form the perfect square trinomial and immediately subtracted so that the value of the equation was not altered. Similarly for the y terms.

Now substitute squared binomials for the perfect square trinomials and simplify the constant terms.

(x - 7/2)^2 + (y + 1)^2 = 7 + 49/4 + 1 = 32/4 + 49/4 = 81/4 = (9/2)^2

This is a circle with center at the point (7/2, -1) and radius 9/2.