Point #J#, located at #(3, 8),# lies on a circle with center #(6, 4)#. What is the equation of the circle?

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Answer 1 · 2017-09-12T18:49:35

# x^2+y^2-12x-8y+27=0.#

Let, #C=C(6,4)# be the Centre of the Circle, and, #r,# the

Radius.

As the pt. #J(3,8)# lies on the circle, we know from Geometry,

#r^2=JC^2=(6-3)^2+(4-8)^2=9+16=25.#

Clearly, the eqn. of the circle is given by,

#(x-6)^2+(y-4)^2=r^2=25, i.e., #

# x^2+y^2-12x-8y+27=0.#