Question

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

x² + y²- 6x - 12y +29=0

Answer 1

Answer: Center: (3,2) Radius: 5

Step-by-step explanation:#x^2 - 6x +y^2 - 4y = 12

Then take 1/2 of the 'b' term for both quadratic expressions, square those values and add them to both sides.

#x^2 -6x + 9 + y^2 - 4y + 4 = 12 + 9 + 4

(x - 3)^2 + (y -2)^2 = 25

Circle centered at (3,2) with radius = 5

Answer 2

radius 4

center (3,6)

Explanation:

(x - h)^2 + (y - k)^2 = r^2

coordinates of the center (h, k) and the radius is (r)

x² + y²- 6x - 12y +29=0

x² - 6x + y²- 12y +29=0

(x² - 6x) + (y²- 12y) +29=0

complete the square

(x² - 6x) + 9 + (y²- 12y) +36 +29= + 9 +36

(x² - 6x + 9) + (y²- 12y + 36) = + 9 +36 -29

(x-3)^2 + (y-6)^2 = 16

(x - h)^2 + (y - k)^2 = r^2

coordinates of the center (h, k) and the radius is (r)

center (3,6)

radius 4

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