Question

Determine the center and radius of the following circle equation:
x2 + y2 – 10x + 8y + 40 = 0

Answer 1

Center (5,-4) Radius = 1

Explanation:

Put it in the form of the standard circle equation after dividing everything by 2. (-10x/2=-5x) (8y/2=4y) Since the radius wasn't given, it's just going to be a 1.

(x-5)^2+y(-(-4))^2 = 1^2

Hence, center (5, -4) and radius 1.

Answer 2

The center of this circle is (5, -4) and the radius is 1.

Explanation:

First regroup these terms according to x and y :

x^2 - 10x + y^2 + 8y = -40

Next, complete the square for x^2 - 10x: x^2 - 10x + 5^2 - 5^2.

and the same for y^2 + 8y: y^2 + 8y + 16 - 16

Substituting these results into x^2 - 10x + y^2 + 8y = -40, we get:

x^2 - 10x + 5^2 - 5^2 y^2 + 8y + 16 - 16 = -40.

Next, rewrite x^2 - 10x + 25 and y^2 + 8y + 16 as squares of binomials:

Then x^2 - 10x + 5^2 - 5^2 y^2 + 8y + 16 - 16 = -40 becomes:

(x - 5)^2 + (y + 4)^2 - 25 - 16 = -40, or:

(x - 5)^2 + (y + 4)^2 = 1

This equation has the form (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center and r is the radius. Matching like terms, we get h = 5, k = -4 and r = 1.

The center of this circle is (5, -4) and the radius is 1.