A circle in the xy -plane has its center at - 5 , 2 and has a radius of 9

Question

A circle in the xy -plane has its center at - 5 , 2 and has a radius of 9 . An equation of this circle is x 2 + y 2 + a x + b y + c = 0 , where a , b , and c are constants. What is the value of c ?

Accepted Answer

The correct answer is - 52 . The equation of a circle in the xy -plane with its center at h , k and a radius of r can be written in the form x - h 2 + y - k 2 = r 2 . It's given that a circle in the xy -plane has its center at - 5 , 2 and has a radius of 9 . Substituting - 5 for h , 2 for k , and 9 for r in the equation x - h 2 + y - k 2 = r 2 yields x - - 5 2 + y - 2 2 = 9 2 , or x + 5 2 + y - 2 2 = 81 . It's also given that an equation of this circle is x 2 + y 2 + a x + b y + c = 0 , where a , b , and c are constants. Therefore, x + 5 2 + y - 2 2 = 81 can be rewritten in the form x 2 + y 2 + a x + b y + c = 0 . The equation x + 5 2 + y - 2 2 = 81 , or x + 5 x + 5 + y - 2 y - 2 = 81 , can be rewritten as x 2 + 5 x + 5 x + 25 + y 2 - 2 y - 2 y + 4 = 81 . Combining like terms on the left-hand side of this equation yields x 2 + y 2 + 10 x - 4 y + 29 = 81 . Subtracting 81 from both sides of this equation yields x 2 + y 2 + 10 x - 4 y - 52 = 0 , which is equivalent to x 2 + y 2 + 10 x + - 4 y + - 52 = 0 . This equation is in the form x 2 + y 2 + a x + b y + c = 0 . Therefore, the value of c is - 52 .

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