How do you find the equation of the circle with center at (4,-6) and passes through point (7,-2)?

1 Answer
Shwetank Mauria
Mar 2, 2017

Equation of a circle is #(x-4)^2+(y+6)^2=25# or #x^2+y^2-8x+12y+27=0#

Explanation:

Equation of a circle with center at #(h,k)# is

#(x-h)^2+(y-k)^2=r^2#, where #r# is its radius.

Hence equation of a circle with center at #(4,-6)# is

#(x-4)^2+(y-(-6))^2=r^2# or #(x-4)^2+(y+6)^2=r^2#

As it passes through #(7,-2)#, we will have

#(7-4)^2+(-2+6)^2=r^2#

or #3^2+4^2=r^2# and #r^2=16+9=25#

Hence, equation of circle is #(x-4)^2+(y+6)^2=25#

or #x^2+y^2-8x+12y+27=0# and its radius is #sqrt25=5#
graph{x^2+y^2-8x+12y+27=0 [-8, 16, -12, 0]}

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