# Find the center and radius for ??????????????

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

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

x2+y2-6x -2y+1 = 0

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.

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