Question

Find the center and radius for x^2+y^2-6x-2y+1=0

Answer 1

The given equation is x²+y²-6x -2y +1 = 0

x²+y²-6x -2y+1 = 0

Add 9 to each side

x²-6x +9+ y²- 2y+1 = 9

(x - 3)²+ (y -1)²= 9

Compare the above equation with given formula (x -h)²+ (y -k)² = r²

Hence (h,k) is center and r is radius

Therefore h = 3, k = 1 and  r = 3

Therefore center (h,k) = ( 3,1) and radius r = 3.

Answer 2

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 + y² - 6x - 2y + 1 = 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 = - 6, so, (half the x coefficient)² = (- 6/2)²= 9.

Here y coefficient = - 2, so, (half the y coefficient)² = (- 2/2)²= 1.

Add 9 and 1 to each side.

x² - 6x + 9 + y² - 2y + 1 = - 1 + 9 + 1

(x - 3)² + (y - 1)² = 9

(x - 3)² + (y - 1)² = (3)²

Compare the equation with standard form of a circle equation.

The center is (3, 1) and

The radius is 3 units.