Question

A circle has a center that falls on the line `y = 2/3x +7` and passes through `(4 ,7 )` and `(1 ,2 )`. What is the equation of the circle?

Answer 1

The equation of the circle is `(x+15/19)^2+(y-123/19)^2=23.2`

Explanation:

Let the center of the circle be `(a,b)` and radius `r`
Then `b=2/3a+7`
The equation of the circle is `(x-a)^2+(y-b)^2=r^2`
Substituting the values of the 2 points
`(4-a)^2+(7-b)^2=r^2`
and `(1-a)^2+(2-b)^2=r^2`
`:.(4-a)^2+(7-b)^2=(1-a)^2+(2-b)^2`
`16-8a+a^2+49-14b+b^2=1-2a+a^2+4-4b+b^2`

`65-8a-14b=5-2a-4b`
`6a+10b=60` `=>` `3a+5b=30`
and the first equation is `3b-2a=21`
Solving for `a` and `b`, we get `(-15/19,123/19)` as the center of the circle
we calculate the radius of the circle
`r^2=(34/19)^2+(85/19)^2=23.21`
`r=4.82`

Equation of the circle
`(x+15/19)^2+(y-123/19)^2=23.2`