How do you find the center and radius of the circle #x^2-8x+y^2-12y= -51#?

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Answer 1 · 2016-11-01T19:04:24

The center is the point #(4, 6)# and the radius, #r = 1#

Add #k^2# and #h^2# to both sides:

#x^2 - 8x + h^2 + y^2 - 12y + k^2 = h^2 + k^2 - 51#

Use the pattern #(x - h)^2 = x^2 - 2hx + h^2# to find the value of #h# and #h^2#:

#x^2 - 2hx + h^2 = x^2 - 8x + h^2#

#-2hx = -8x#

#h = 4#

#h^2 = 16#

Write the left side as a perfect square and substitute 16 for #h^2# on the right:

#(x - 4)^2 + y^2 - 12y + k^2 = 16 + k^2 - 51#

Use the pattern #(y - k)^2 = y^2 - 2ky + k^2# to find the value of #k# and #k^2#:

#y^2 - 2ky + k^2 = y^2 - 12y + k^2#

#-2ky=-12y#

#k = 6#

#k^2 = 36#

Write the left side as a perfect square and substitute 36 for #k^2# on the right:

#(x - 4)^2 + (y - 6)^2 = 16 + 36 - 51#

#(x - 4)^2 + (y - 6)^2 = 1#