Question

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

Answer 1

Center = (5, -6)

Radius = 7

Explanation:

To find the center and the radius of the circle represented by the given equation, rewrite the equation in standard form by completing the square.

To complete the square, begin by moving the constant to the right side of the equation and collecting like terms on the left side of the equation:

`x^2-10x+y^2+12y=-12`

Add the square of half the coefficient of the term in x and the term in y to both sides of the equation:

`x^2-10x+\left((-10)/(2)\right)^2+y^2+12y+\left((12)/(2)\right)^2=-12+\left((-10)/(2)\right)^2+\left((12)/(2)\right)^2`

Simplify:

`x^2-10x+(-5)^2+y^2+12y+(6)^2=-12+(-5)^2+(6)^2`

`x^2-10x+25+y^2+12y+36=-12+25+36`

`x^2-10x+25+y^2+12y+36=49`

Factor the perfect square trinomials on the left side:

`(x-5)^2+(y+6)^2=49`

The standard equation of a circle is:

`\boxed{(x-h)^2+(y-k)^2=r^2}`

where:

• (h, k) is the center.
• r is the radius.

Comparing this with the rewritten given equation, we get

• 
  `h = 5`
• 
  `k = -6`
• 
  `r^2 = 49 \implies r=7`

Therefore, the center of the circle is (5, -6) and its radius is r = 7.

Answer 2

centre = (5, - 6 ) , radius = 7

Explanation:

the equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k ) are the coordinates of the centre and r is the radius

given

x² + y² - 10x + 12y + 12 = 0 ( subtract 12 from both sides )

x² + y² - 10x + 12y = - 12 ( collect terms in x/ y )

x² - 10x + y² + 12y = - 12

using the method of completing the square

add ( half the coefficient of the x/ y terms )² to both sides

x² + 2(- 5)x + 25 + y² + 2(6)y + 36 = - 12 + 25 + 36

(x - 5)² + (y + 6)² = 49 = 7² ← in standard form

with centre (5, - 6 ) and radius = 7