How do you graph #x^2+y^2-8x+6y+16=0#?

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Answer 1 · 2016-11-15T00:30:57

Set your compass to a radius of 3, put the centerpoint at #(4, -3)#, and draw a circle.

This is a circle. To find the center and the radius, we must complete the squares for both x terms and y terms:

Add #h^2 + k^2 -16 # to both sides:

#x^2 - 8x + h^2 + y^2 + 6y + k^2 = h^2 + k^2 - 16#

Use the linear x term to find the value of h by setting it equal to -2xh:

#-2hx = -8x#

#h = 4#

Substitute #(x - 4)^2# for the x terms on the left and 16 for #h^2# on the right:

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

Use the linear y term to find the value of k by setting it equal to -2yk:

#-2yk = 6y#

#k = -3#

Substitute #(y - -3)^2# for the y terms on the left and 9 for #k^2# on the right:

#(x - 4)^2 + (y - -3)^2 = 16 + 9 - 16 = 9 = 3^2#

The final equation is:

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

This is a circle with a center of #(4, -3)# and a radius of 3.