How do you find the coordinates of the center of a circle whose equation is #x^2-14x+y^2+6y+54=0#?

1 Answer
Jan 28, 2016

Answer:

Complete squares and rearrange into the standard form of the equation of a circle to find centre #(h, k) = (7, -3)# and radius #2#

Explanation:

Complete the squares for both #x# and #y#:

#0 = x^2-14x+y^2+6y+54#

#= (x^2-14x+49) + (y^2+6y+9) + (54-49-9)#

#= (x-7)^2+(y+3)^2-4#

Add #4# to both ends and slightly re-express to get:

#(x-7)^2 + (y-(-3))^2 = 2^2#

This is in the standard form of the equation of a circle:

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

with centre #(h, k) = (7, -3)# and radius #r=2#.

graph{(x^2-14x+y^2+6y+54)((x-7)^2+(y+3)^2-0.01) = 0 [-6.79, 13.21, -6.16, 3.84]}