How do you find the general form of the equation of a circle given the center at the point (1, 0) and has the point (-2, 3)?

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Answer 1 · 2016-09-08T05:44:55

#x^2+y^2-2x-17=0#

The general form of equation of a circle is

#x^2+y^2+2gx+2fy+c=0#, whose center is #(-g,-f)#.

As the center is #(1,0)#, the equation would be of the form

#x^2+y^2+2(-1)x+2×0×y+c=0# or

#x^2+y^2-2x+c=0#

As it passes through #(-2,3)#, putting these values in the equation we get #c#.

#(-2)^2+3^2-2×(-2)+c=0# or

#4+9+4+c=0# i.e. #c=-17#.

Hence equation of circle is

#x^2+y^2-2x-17=0#