How do you find the center and radius for #x^2 - 3x + y^2 - 3y - 20 = 0#?

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Answer 1 · 2018-07-28T06:42:57

The center is #=(3/2,3/2)# and the radius #r=7sqrt2/2#

The standard equation of a circle is

#(x-a)^2+(y-b)^2=r^2#

where

The center is #(a,b)# and the radius is #=r#

Here, we have

#x^2-3x+y^2-3y-20=0#

Complete the square for #x# and #y#

#x^2-3x+9/4+y^2-3y+9/4=20+9/4+9/4#

#(x-3/2)^2+(y-3/2)^2=98/4=(7sqrt2/2)^2#

The center is #=(3/2,3/2)# and the radius #r=7sqrt2/2#

graph{x^2-3x+y^2-3y-20=0 [-16.35, 15.68, -4.36, 11.66]}