How do you find the center and radius of the circle with this equation #x^2 + y^2 – 8x – 20y + 80 = 0#?

1 Answer
Trevor Ryan.
Jan 18, 2016

The centre is #(4,10)# and the radius is #sqrt98#.

Explanation:

You would have to complete the square twice, once for x and once for y, in order to write in in standard form #(x-a)^2+(y-b)^2=r^2#, from which the centre would be #(a,b)# and the radius #r#.

Completing the square to write it in this form yields :

#x^2-8x+((-8)/2)^2-((-8)/2)^2+y^2-20y+((-20)/2)^2-((-20)/2)^2+80=0#

#therefore(x-4)^2+(y-10)^2-16-100+80=0#

#therefore(x-4)^2+(y-10)^2=98#.

Hence the centre is #(4,10)# and the radius is #sqrt98#.

Related questions
See all questions in Identify Critical Points
Impact of this question
2097 views around the world
You can reuse this answer
Creative Commons License