Question

Find the center and radius of the circle represented by the equation below.

x^2+y^2-8x-20y+52=0

Answer 1

Explanation:

x² + y² - 8x - 20y + 52 = 0

the standard circle equation is

(x - h)² + (y - k)² = r²

where (h, k) is the center of the circle, and r is the radius.

now, we need to find the squared terms in the original equation.

x² and y² are clearly from the (x - h)² and (y - k)².

and then

(x - h)² = x² - 2hx + h²

-2hx = -8x

2h = 8

h = 4

(y - k)² = y² - 2ky + k²

-2ky = -20y

2k = 20

k = 10

h² + k² = 16 + 100 = 116

116 - r² = 52

r² = 64

r = 8

so, the center is (4, 10), the radius is 8.

Answer 2

Explanation:

To find the center and radius of the circle represented by the equation:

x^2 + y^2 - 8x - 20y + 52 = 0

We can start by completing the square for both x and y terms as follows:

(x^2 - 8x) + (y^2 - 20y) + 52 = 0

(x^2 - 8x + 16) + (y^2 - 20y + 100) + 52 = 16 + 100

(x - 4)^2 + (y - 10)^2 = 64

Thus, the equation can be rewritten in the standard form of the circle as:

(x - 4)^2 + (y - 10)^2 = 8^2

Therefore, the center of the circle is (4, 10), and its radius is 8 units.