SOLUTION: A circle is defined by the following equation: 2x^2+2y^2+28x+80=0 What is the radius?

Question 82851: A circle is defined by the following equation:
2x^2+2y^2+28x+80=0
What is the radius?
Answer by funmath(2933) (Show Source): You can put this solution on YOUR website!
A circle is defined by the following equation:
2x^2+2y^2+28x+80=0
What is the radius?
The standard form of the equation of a circle is: , where the center is (h,k) and the radius is r.
First divide everything by 2 because x^2 and y^2 have to have 1 as their coefficients.

Now get the x's next to each other the y's next to each other and the constants on the other side of the = sign:
x^2+14x+y^2+40-40=0-40
x^2+14x+y^2=-40
Now complete the square for the x terms. Add 1/2 the coefficient of x squared to both sides of the =sign.
x^2+14x+(14/2)^2+y^2=-40+(14/2)^2
x^2+14x+(7)^2+y^2=-40+(7)^2
x^2+14x+49+y^2=-40+49
(x^2+14x+49)+y^2=9
(x+7)^2+(y-0)^2=3^2
The radius is 3.
You didn't ask for the center, but it's (-7,0)
Happy Calculating!!!!
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