Question

A circle has a center that falls on the line `y = 5/3x +1` and passes through `(5 ,2 )` and `(3 ,2 )`. What is the equation of the circle?

Answer 1

`3x^2+3y^2-24x-46y+125=0.`

Explanation:

Let the given points be `A(5,2), B(3,2).`

Let us observe that the y-co-ordinates of A & B are same, hence, `AB` a is a horizontal chord.

We know that the centre of a circle lies on the perpendicular bisector (p.b.) of a chord. Chord `AB` is horizontal, so, its p.b. is a vertical line thro. the midpoint of `AB`, i.e., `x=(5+3)/2.` The cetre is on `x=4...... (i)`

By what is given, centre is also on the line `y=(5/3)x+1. .......(ii)`

Solving `(i)` and `(ii)`, `y=23/3.`

So, we get the centre `(4,23/3).`

Next goal is to find radius r of the reqd. cicle. We note that

`CA=CB=r.`

We use `CB^2` = `r^2= (4-3)^2+(23/3-2)^2=1+289/9.`

Hence, the reqd. eqn.of the circle is `(x-4)^2+(y-23/3)^2=r^2=(4-3)^2+(23/3-2)^2.`
`:. (x-4)^2+(y-23/3)^2`-- `(4-3)^2 - (23/3-2)^2.`
`:.(x-4)^2- (4-3)^2+(y-23/3)^2- (23/3-2)^2=0..`
`:.(x-4+4-3)(x-4-4+3)(y-23/3+23/3-2)(y-23/3-23/3+2)=0.`
`:. (x-3)(x-5)+(y-2)(y-40/3)=0.`
`:.3(x^2-8x+15)+(y-2)(3y-40)=0.`
`:. 3x^2-24x+45+3y^2-46y+80=0.`
`:. 3x^2+3y^2-24x-46y+125=0.`