What is the center and radius of the circle described by this equation #x^ { 2} - 16x + y ^ { 2} - 12y = - 51#?

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Answer 1 · 2017-06-15T18:25:45

This is the equation of a circle, center #(8,6)# and the radius is #=7#

We need

#a^2-2ab+b^2=(a-b)^2#

The equation of a circle, center #(a,b)# and radius #=r# is

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

Our equation is

#x^2-16x+y^2-12y=-51#

We complete the squares by adding half the coefficient of #x# and #y# to the square on both sides of the equation

#x^2-16+64+y^2-12y+36=-51+64+36#

We factorise the LHS

#(x-8)^2+(y-6)^2=49=7^2#

This is the equation of a circle, center #(8,6)# and the radius is #=7#

graph{(x^2-16x+y^2-12y+51)=0 [-7.42, 21.06, -1.37, 12.88]}