How do you graph #x^2+y^2-2x-4y-20=0#?

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Answer 1 · 2016-07-28T06:58:33

This is a circle with centre #(1, 2)# and radius #5#.

Complete the square for both #x# and #y# ...

#0 = x^2+y^2-2x-4y-20#

#= color(blue)(x^2-2x+1)+color(green)(y^2-4y+4)-25#

#= (x-1)^2+(y-2)^2-5^2#

Add #5^2# to both ends and transpose to get:

#(x-1)^2+(y-2)^2=5^2#

This is in the form:

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

the equation of a circle with centre #(h, k) = (1, 2)# and radius #r=5#

graph{((x-1)^2+(y-2)^2-25)((x-1)^2+(y-2)^2-0.01) = 0 [-9.04, 10.96, -2.76, 7.24]}