Question

A circle has a center that falls on the line `y = 3/8x +5` and passes through `( 7 ,3 )` and `(2 ,9 )`. What is the equation of the circle?

Answer 1

`16(x-6)^2 + (4y-29)^2 = 305`

Explanation:

The center should be equidistant from `(7,3)` and `(2,9)`. Use the Pythagoras Theorem to find the distance between two points.

`sqrt{(x-7)^2+(y-3)^2} = sqrt{(x-2)^2+(y-9)^2}`

`(x-7)^2 + (y-3)^2 = (x-2)^2 + (y-9)^2`

`(x^2-14x+49) + (y^2-6y+9) = (x^2-4x+4) + (y^2-18y+81)`

`-14x+49 - 6y+9 = -4x+4 -18y+81`

`12y = 10x+27`

The result is an equation of a straight line, that all the points on it are equidistant from `(7,3)` and `(2,9)`. We will find where it intersects with the line `y = 3/8 x + 5` to locate the center of the circle.

`y = 3/8 x + 5`
`12y = 10x+27`

Solve them simultaneously.

`12(3/8 x + 5) = 10x+27`

`9 x + 120 = 20x+54`

`66 = 11x`

`x = 6`

`y = 3/8 (6) + 5`

`= 29/4`

The center of the circle is at `(6,29/4)`. You can verify that it is indeed equidistant from `(7,3)` and `(2,9)`.

`sqrt{(6-7)^2+(29/4-3)^2} = sqrt305/4`

`sqrt{(6-2)^2+(29/4-9)^2} = sqrt305/4`

Bear in mind that the radius is the distance from the center to any point on the circumference. Therefore, the radius is `sqrt305/4`.

For a circle with center at `(a,b)` and radius `r`, the cartesian equation is

`(x-a)^2 + (y-b)^2 = r^2`

Our circle has center at `(6,29/4)` and radius `sqrt305/4`, so its equation is

`(x-6)^2 + (y-29/4)^2 = (sqrt305/4)^2`

`16(x-6)^2 + (4y-29)^2 = 305`