How do you find the equation of the circle with center at (3, 2) and through the point (5, 4)?

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Answer 1 · 2016-06-18T21:31:04

The equation is #x^2+y^2-6x-4y+5=0#.

The generic equation of a circle with center #(c_x, c_y)# and radius #r# is:

#(x-c_x)^2+(y-c_y)=r^2#

in our case

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

What is missing here is the radius, but we can find it substituting the point #(5,4)#

#(5-3)^2+(4-2)^2=r^2#

#2^2+2^2=r^2#

#4+4=r^2#

#r=sqrt(8)#

The final equation is

#(x-3)^2+(y-2)^2=8# or, expanding the squares,

#x^2+y^2-6x-4y+5=0#.