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#?

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Answer 1 · 2016-01-08T20:33:06

Rearrange into the form:

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

to identify the centre #(4,3)# and radius #2#.

Complete the squares for #x# and #y#...

#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 #2^2# to both ends and transpose to get:

#(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 #(h, k) = (4, 3)# and radius #r=2#

Notice that I picked out the constant value #16# to complete the square #(x-4)^2 = x^2-8x+16# and #9# to complete the square #(y-3)^2 = y^2-6y+9#.