The equation of a circle is written as x² + 10 + y² - 12y + 1 = 0. What is the equation of this circle in standard form?

Question

[Enter your answer as (x - h)² + (y - k)² = r².]

Answer 1

x² + 10 + y² - 12y + 1 = 0

x² + y² - 12y + 11 = 0

x² + y² - 12y = -11

x² + y² - 12y + 36 = -11+ 36

x² + (y² - 12y + 36) = 25

Therefore the standard form is:

(x-0)² + (y-6)² = 5²

Center is at C(h,k)= (0,6) with a radius of r=5

Answer 2

x² + y² - 12y = -1

Complete the square for y.

Convert to required form.