How do you find the center and radius of the circle #x^2-6x+60+y^2-20y=0#?

Question

1459 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-23T09:55:20

Center of the circle is #(3,10)# and radius is #7#.

If radius of a circle is #r# and its center is #(h,k)#, the equation of circle is

#(x-h)^2+(y-k)^2=r^2#

Hence let us try to convert the given equation in this form

#x^2-6x+60+y^2-20y=0# or

#x^2-2xx3x+3^2+y^2-2xx10y+10^2-3^2-10^2+60=0# or

#(x^2-2xx3x+3^2)+(y^2-2xx10y+10^2)=3^2+10^2-60# or

#(x-3)^2+(y-10)^2=9+100-60=49# or

#(x-3)^2+(y-10)^2=7^2#

Hence center of the circle is #(3,10)# and radius is #7#.
graph{x^2-6x+60+y^2-20y=0 [-17.41, 22.59, -1.36, 18.64]}