Question

The equation of a circle is x2+y2−12x+6y+20=0 .

Answer 1

Answer:

The radius is r=5 units

Step-by-step explanation:

we know that

The equation of the circle in standard form is equal to

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

where

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

we have

$x^{2}+y^{2}-12x+6y+20=0$

Convert to standard form

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

$(x^{2}-12x)+(y^{2}+6y)=-20$

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

$(x^{2}-12x+36)+(y^{2}+6y+9)=-20+36+9$

$(x^{2}-12x+36)+(y^{2}+6y+9)=25$

Rewrite as perfect squares

$(x-6)^{2}+(y+3)^{2}=5^{2}$

therefore

The center is the point (6,-3) and the radius is r=5 units