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Jefferson Simmons

Answered question

2023-03-16

How to find the center and radius of the circle

x^{2} - 12 x + y^{2} + 4 y + 15 = 0

Maverick Carney

Beginner2023-03-17Added 7 answers

We rearrange the equation and finish the squares.

x^{2} - 12 x + y^{2} + 4 y = - 15

x^{2} - 12 x + 36 + y^{2} + 4 y + 4 = - 15 + 36 + 4

And now we factorise

{(x - 6)}^{2} + {(y + 2)}^{2} = 25 = 5^{2}

We compare this equation to the standard equation of a circle

{(x - a)}^{2} + {(y - b)}^{2} = r^{2}

The centre is

(a, b)

and the radius

= r

In our case,
The center is

(6, - 2)

and the radius

= 5

graph{x^2-12x+y^2+4y+15=0 [-6.22, 16.275, -6.95, 4.3]}

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