Search 75,663 tutors
FIND TUTORS
Ask a question
0 0

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

Tutors, please sign in to answer this question.

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.