Question

What is the equation of a circle of radius 1, which is also tangent to the line 3x-4y+1=0 at the y-axis value = 1 ?

Answer 1

There are two possible circles, either side of the line:

`(x-2/5)^2 + (y-9/5)^2 - 1 = 0`

`(x-8/5)^2 + (y - 1/5)^2 - 1 = 0`

Explanation:

When `y=1` the equation becomes `3x-4+1 = 0`, hence `x=1`.

So the circle is tangential to the line at `(1, 1)`.

Starting with `3x-4y+1 = 0`, add `4y` to both sides to get:

`4y = 3x+1`

Divide both sides by `4` to get:

`y = 3/4x+1/4`

The slope of this line is `m = 3/4`, so the slope of the radius of the tangential circle passing through the point of contact will be `-1/m = -4/3`.

So this perpendicular line will have equation:

`y = -4/3x + 7/3` in order that it passes through `(1,1)`

The centre of the tangential circle lies at distance `1` from `(1,1)` along this line.

That give two possibilities:

`(1 - 3/5, 1 + 4/5) = (2/5, 9/5)`

and

`(1 + 3/5, 1 - 4/5) = (8/5, 1/5)`

...using the right angled property of the 3, 4, 5 triangle.

The corresponding circles have equations:

`(x-2/5)^2 + (y-9/5)^2 - 1 = 0`

`(x-8/5)^2 + (y - 1/5)^2 - 1 = 0`

graph{(3x-4y+1)((x-2/5)^2 + (y-9/5)^2 - 1)((x-8/5)^2 + (y - 1/5)^2 - 1) = 0 [-3.52, 6.48, -1.28, 3.72]}