How do you write the standard equation of the circle the given center that passes with through the given point: center (5,9); point (2, 9)?

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Answer 1 · 2017-07-06T12:55:52

$(x-5)^2+(y-9)^2=3^2, or, x^2+y^2-10x-18y-97=0.$

The Centre $C(h,k)$ of the Circle is $C(h,k)=C(5,9)$ and the

circle passes through a point $P(2,9).$

Hence, using the Distance Formula, the Radius

$r=CP=sqrt{(5-2)^2+(9-9)^2}=3.$

We know that, the Eqn. of the circle having centre at $(h,k)$ and

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

Therefore, the Reqd. Eqn. is,

$(x-5)^2+(y-9)^2=3^2, or, x^2+y^2-10x-18y-97=0.$