Question

PLEASE HELP Identify the radius and center x^2 +y^2 -6x -2y + 1 = 0

Answer 1

Answer:

The center is the point (3,1) and the radius is 3 units

Step-by-step explanation:

we know that

The equation of a circle in standard form is equal to

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

we have

$x^{2}+y^{2}-6x-2y+1=0$

Convert to standard form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2}-6x)+(y^{2}-2y)=-1$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^{2}-6x+9)+(y^{2}-2y+1)=-1+9+1$

$(x^{2}-6x+9)+(y^{2}-2y+1)=9$

Rewrite as perfect squares

$(x-3)^{2}+(y-1)^{2}=9$

The center is the point (3,1) and the radius is 3 units