How do you find the center and radius of the circle given #x^2+y^2-18x-18y+53=0#?

1 Answer
Oct 13, 2016

Answer:

Complete the square for the x and y terms to find the center #=(9,9)# and the radius #r=sqrt109#.

Explanation:

#x^2+y^2-18x-18y+53=0#

Use a method called "Completing the Square".

1) Group the x terms and y terms. Move the constant term to the right side of the equation by subtracting 53 from both sides.

#(x^2color(limegreen)(-18)xcolor(white)(aaa))+(y^2color(magenta)(-18)ycolor(white)(aaa))=-53#

2) Divide the coefficient of the x term by 2 and then square it.

#color(limegreen)(-18)/2= -9color(white)(aaa)(-9)^2 =color(red)(81)#

3) Add the result to both sides.

#(x^2color(limegreen)(-18)x+color(red)(81))+(y^2color(magenta)(-18)ycolor(white)(aaa))=-53+color(red)(81)#

4) Divide the coefficient of the y term by 2 and then square it.

#color(magenta)(-18)/2=-9color(white)(aaa)(-9)^2=color(blue)(81)#

5) Add the result to both sides.

#(x^2-18x+color(red)(81))+(y^2-18y+color(blue)(81))=-53+color(red)(81)+color(blue)(81)#

Factor each set of parentheses. Note that the #-9# in factored form is the same number you got when dividing the coefficient of the middle term.

#(x-9)(x-9)+(y-9)(y-9)=109#

Rewrite as the square of a binomial.

#(x-9)^2+(y-9)^2=109#

Compare this equation to the equation of a circle
#(x-h)^2+(y-k)^2=r^2#
where #(h,k)# is the center and #r# is the radius.

In this example, the center #(h,k)=(9,9)# and the radius #r=sqrt109#