# How do you find the center and radius of the circle given by the equation #x^2+y^2-8 x- 6 y +21=0#?

##### 1 Answer

Jan 8, 2016

Rearrange into the form:

#(x-4)^2+(y-3)^2 = 2^2#

to identify the centre

#### Explanation:

Complete the squares for

#0 = x^2+y^2-8x-6y+21#

#=(x^2-8x+16) + (y^2 -6y + 9) - 4#

#=(x-4)^2+(y-3)^2-2^2#

Add

#(x-4)^2+(y-3)^2 = 2^2#

This is in the form:

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

which is the equation of a circle with centre

Notice that I picked out the constant value