Question

A circle has a center that falls on the line `y = 2/9x +8` and passes through `( 2 ,5 )` and `(1 ,7 )`. What is the equation of the circle?

Answer 1

Explanation:

Let the coordinate of the centre of the circle be (a,b) and radius is r.
Then its equation will be
`(x-a)^2+(y-b)^2=r^2`

The circle passes through (2,5) and (1,7) show these points will satisfy the equation of the circle, hence

`(2-a)^2+(5-b)^2=r^2.......(1)`
`(1-a)^2+(7-b)^2=r^2.......(2)`
Subtracting (2) from (1) we have

`(2-a)^2-(1-a)^2+(5-b)^2-(7-b)^2=0`

`=>(2-a-1+a)(3-2a)+(5-b-7+b)(12-2b)=0`
`=>3-2a-24+4b=0`
`=>2a-4b=-21......(3)`

Again The center (a,b) lies on the given line `y=2/9x+8` So we can write `b=2/9a+8......(4)`

From (3) and (4)
we have
`=>2a-4(2/9a+8)=-21`

`=>2a-8/9a-32=-21`
`=>2a-8/9a=11`
`=>10a=99`
`:,a=9.9`
From (4)
`b=2/9a+8`
`=>b=2/9xx9.9+8=10.2`
From eq (1)
`(2-9.9)^2+(5-10.2)^2=r^2.......(1)`
`=>r^2=62.41+27.04=89.45`

Equation of the Circle
`(x-9.9)^2+(y-10.2)^2=89.45^2`