How do you write the equation of the circle with the given center at (-2,4) and passes through the point at (1,-7)?

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Answer 1 · 2017-03-02T09:52:54

Equation of the circle is #(x+2)^2+(y-4)^2=121# or #x^2+y^2+4x-8y-101=0#

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 #(-2,4)# is

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

As it passes through #(1,-7)#, we will have

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

or #0^2+11^2=r^2# and #r^2=0+121=121#

Hence, equation of circle is #(x+2)^2+(y-4)^2=121#

or #x^2+y^2+4x-8y-101=0# and its radius is #sqrt121=11#
graph{x^2+y^2+4x-8y-101=0 [-28, 24, -10, 16]}