What is the equation, center, and radius of the circle that passes through #(4,3), (0,-5),# and #(1,2)#?

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Answer 1 · 2016-09-21T16:55:49

#(x-4)^2+(y+2)^2=25#

Radius is #5#

Center is #(4,-2)#

A equation for a circle takes the general form

#x^2+y^2+Ax+By+C=0#

We are given #3# points that are on the circle so we can write
an equation for each point

#4^2+3^2+4A+3B+C=0#

#0^2+(-5)^2+0A+(-5)B+C=0#

#1^2+2^2+1A+2B+C=0#

Now we simplify where we can

#16+9+4A+3B+C=0#

#25-5B+C=0#

#5+1A+2B+C=0#

Now combine like terms

#25+4A+3B+C=0#

#25-5B+C=0#

#5+1A+2B+C=0#

Now rewrite by subtracting the constant term from both sides.

#4A+3B+C=-25#

#-5B+C=-25#

#1A+2B+C=-5#

I will assume that you know how to solve a system of equations

Doing so you get

#A=-8#

#B=4#

#C=-5#

So our general form for the equation of the circle is

#x^2+y^2-8x+4y-5=0#

Add #5# to both sides and rewrite

#x^2-8x+y^2+4y=5#

Now complete the squares as follows

#x^2-8x+16+y^2+4y+4=5+16+4#

#(x-4)^2+(y+2)^2=25#