Question

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

Answer 1

`(x-16/9)^2 + (y-65/9)^2 = 404/81`

Explanation:

The centre of the circle must lie on the perpendicular bisector of the line segment joining `(4, 7)` and `(2, 5)`.

Their midpoint is:

`((4+2)/2, (7+5)/2) = (3, 6)`

and the slope of the line segment joining them is:

`(5-7)/(2-4) = (-2)/(-2) = 1`

Hence the slope of the perpendicular bisector is:

`-1/color(blue)(1) = -1`

Hence the equation of the perpendicular bisector can be written:

`y - 6 = -1(x - 3)`

which simplifies to:

`y = 9-x`

This will intersect the given line when:

`9-x = y = 5/4x+5`

Multiply both ends by `4` to get:

`36-4x=5x+20`

Add `4x-20` to both sides to get:

`16 = 9x`

Hence:

`x = 16/9`

and:

`y = 9-x = 9-16/9 = 65/9`

So the centre of the circle is at: `(16/9, 65/9)`

So its equation can be written in the form:

`(x-16/9)^2 + (y-65/9)^2 = r^2`

where the radius `r` is yet to be determined.

Since this passes through `(4, 7)`, those values of `x` and `y` must satisfy the equation.

So we find:

`r^2 = (color(blue)(4)-16/9)^2 + (color(blue)(7)-65/9)^2`

`color(white)(r^2) = (20/9)^2 + (-2/9)^2`

`color(white)(r^2) = 404/81`

So the equation of the circle may be written:

`(x-16/9)^2 + (y-65/9)^2 = 404/81`

graph{(y-x-3)((x-4)^2+(y-7)^2-0.006)((x-2)^2+(y-5)^2-0.006)((x-16/9)^2 + (y-65/9)^2 - 404/81)((x-16/9)^2 + (y-65/9)^2 - 0.006)(y+x-9)(y-5/4x-5)=0 [-3.63, 6.37, 4.6, 9.6]}