Find the center and radius of the circle. y2 + x2 − 10x + 12y + 45 = 0 center (x, y) = radius

Question

Find the center and radius of the circle.

y² + x² − 10x + 12y + 45 = 0

center	(x, y) =
radius

Answer 1

You have to put the equation into the standard format for a circle:

(x + a)² + (y + b)² = r²

^{First, group the terms consistent with the format:}

y² + x² − 10x + 12y + 45 = 0

y² + 12y + x² - 10x = -45

Next, recognize that the x and y terms must be perfect squares:

y² + 12y + 36 + x² - 10x + 25 = -45 + 36 + 25 <------added constants to both sides

Finally,simplify into the standard format:

(y + 6)² + (x - 5)² = 16

This gives you the center and the radius. For the center, find the x or y value that causes each term in parentheses to be zero. The radius is simply √16 = 4

Answer 2

First, we must know that the equation for a circle with center at (h,k) and radius of r is

(x-h)² + (y-k)² = r²

The equation in this problem is y² + x² − 10x + 12y + 45 = 0

To get into the proper form, we need to complete the square

Gather the x-terms and y-terms separately and move the constant term of 45 to the right side of the equation.

x² -10x + y² +12y = -45

Take half the coefficient of the x term, square it and add to both sides of the equation. Take half the coefficient of the y term, square it and add to both sides of the equation.

Half of -10 is -5, squared is 25. Half of 12 is 6, squared is 36. When adding 25 and 36 to both sides, we get

x² - 10x + 25 + y² + 12y + 36 = -45 + 25 + 36

Notice that we know have two perfect square trinomials, one in terms of x and one in terms of y. Factoring each and adding the terms on the right side of the equals gives us

(x - 5)² + (y + 6)² = 16

Therefore (h,k) = (5, -6) and r² = 16, with r = 4

The center is (5,-6) with a radius of r