How do you find the center and radius of the circle #x^2-12x+y^2+4y+15=0#?

Question

3583 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-18T06:42:24

The center is #(6,-2)# and the radius #=5#

We complete the squares and rearrange the equation

#x^2-12x+y^2+4y=-15#

#x^2-12x+36+y^2+4y+4=-15+36+4#

And now we factorise

#(x-6)^2+(y+2)^2=25=5^2#

We compare this equation to the standard equation of a circle

#(x-a)^2+(y-b)^2=r^2#

The centre is #(a,b)# and the radius #=r#

In our case,

The center is #(6,-2)# and the radius #=5#

graph{x^2-12x+y^2+4y+15=0 [-6.22, 16.275, -6.95, 4.3]}