what are the coordinates of the center of circle shown below, x^2+y^2-12x-6y+9=0?

Question

precalculus

Answer 1

The standard form of the circle equation is ( x - h )² + ( y - k )² = r², where, (h, k) is the center of the circle, and r is the radius.

The equation is x² - 12x + y² - 6y + 9 = 0.

Write the equation in standard form of a circle.

To change the expression into a perfect square  add (half the x coefficient)² and add (half the y coefficient)²to each side of the expression.

Here, x coefficient = -12, so, (half the x coefficient)² = (-12/2)²= 36.

Here, y coefficient = -6, so, (half the y coefficient)² = (-6/2)²= 9.

Add 36 and 9 to each side.

x² - 12x + 36 + y² - 6y + 9 + 9 = 0 + 36 + 9

(x - 6)² + (y - 3)² = 36 + 9-9= 36

(x - 6)² + (y - 3)²  = (6)² .

Compare the equation with standard form of a circle equation.

The center (h, k) is (6, 3), and

The radius (r) is 6 units.