What are the coordinates of the center of the circle #x^2 + y^2 -8x -10y -8=0#?

Question

4029 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-17T22:21:46

#(4, 5)#

Complete the squares for #x# and #y# as follows:

#0 = x^2+y^2-8x-10y-8#

#= color(blue)(x^2-8x+16)+color(green)(y^2-10y+25)-49#

#= color(blue)((x-4)^2)+color(green)((y-5)^2)-7^2#

Add #7^2# to both ends and transpose to get:

#(x-4)^2+(y-5)^2 = 7^2#

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

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

with center #(h, k) = (4, 5)# and radius #r=7#.

graph{((x-4)^2+(y-5)^2-7^2)((x-4)^2+(y-5)^2-0.06)=0 [-16.67, 23.33, -5.2, 14.8]}