Question

A circle has a center that falls on the line `y = 5/4x +4` and passes through `( 4 ,7 )` and `(2 ,5 )`. What is the equation of the circle?

Answer 1

`(x-20/9)^2 + (y - 61/9)^2 = 260/81`

Explanation:

Let's write the center line `4y=5x+16` and let `x=4c` so the center is `(4c, 5c + 4)` eliminating the messy fractions. Let's call the squared radius `k`. The equation for the circle is thus

`(x - 4c)^2 + (y - (5c + 4))^2 = k`

Our job is to find `c` and `k`; our two points give us two equations:

`(4 -4c)^2 + (7 - (5 c + 4))^2 = k`

`(2 - 4c)^2 + (5 - (5 c + 4))^2 = k`

We expand and subtract, which should give us an equation for `c.`

`16 - 32 c + 16c^2 + 9 - 30c + 25c^2 = k`

`4 - 16 c + 16c^2 + 1 -10 c + 25c^2 = k`

`20 -36 c = 0`

`c = 20/36= 5/9`

`(x - 4(5/9))^2 + (y - (5(5/9) + 4))^2 = k`

`(x-20/9)^2 + (y - 61/9)^2 = k`

`k =(2-20/9)^2 + (5 - 61/9)^2 = 260/81`

Our equation is

`(x-20/9)^2 + (y - 61/9)^2 = 260/81`

I'll plot after I post. Plot:

`0 = ( (x-20/9)^2 + (y - 61/9)^2 -260/81)( 5/4 x+4 -y)((x-4)^2+(y-7)^2-.1^2)( (x-2)^2+(y-5)^2-.1^2) ((x-20/9)^2 + (y - 61/9)^2 - .1^2)`

graph{0 = ( (x-20/9)^2 + (y - 61/9)^2 -260/81)( 5/4 x+4 -y)((x-4)^2+(y-7)^2-.1^2)( (x-2)^2+(y-5)^2-.1^2) ((x-20/9)^2 + (y - 61/9)^2 - .1^2) [-2.885, 8.365, 3.577, 9.2]}