Question

How do you find the equation of a circle passes through the points (2,0) and (8,0) and has the y-axis as a tangent?

Answer 1

There are two possible circles:

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

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

Explanation:

The centre of the circle must be on the perpendicular bisector of these two points, namely the line `x=5`.

Since it is tangential to the `y` axis, the radius must be `5` too.

The midpoint of `(2, 0)` and `(8, 0)` is `(5, 0)` which is at distance `3` from each of them.

So the centre of the circle must form `3-4-5` triangles with the points `(2, 0)`, `(5, 0)` and `(5, 0)`, `(8, 0)`. Thus the centre is at `(5, 4)` or `(5, -4)`.

The equation of a circle with centre `(h, k)` and radius `r` can be written:

`(x-h)^2+(y-k)^2=r^2`

Hence there are two possible circles:

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

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

graph{((x-5)^2+(y-4)^2-25)((x-5)^2+(y+4)^2-25)((x-2)^2+y^2-0.1)((x-8)^2+y^2-0.1)((x-5)^2+(y-4)^2-0.06)((x-5)^2+(y+4)^2-0.06) = 0 [-16.25, 23.75, -9.6, 10.4]}