Question

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)?

Answer 1

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

Explanation:

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.`