How do you write the equation of the circle with Center at (4, 4); passing through (7, 14)?

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Answer 1 · 2016-10-19T10:32:23

Equation of circle is #x^2+y^2-8x-8y-77=0#

As the circle has center at #(4,4)# and passes through #(7,14)#,

distance between a point #(x,y)# and #(4,4)# and between #(7,14)# and #(4,4)# will be equal. Hence,

#sqrt((x-4)^2+(y-4)^2)=sqrt((7-4)^2+(14-4)^2)#

or squaring #(x-4)^2+(y-4)^2=(7-4)^2+(14-4)^2#

or #x^2-8x+16+y^2-8y+16=3^2+10^2#

or #x^2+y^2-8x-8y+32-9-100=0#

or #x^2+y^2-8x-8y-77=0#