Question

How do you find the equation of the circle with a radius of `1` inscribed in the parabola `y=x^2`?

Answer 1

`x^2+(y-5/4)^2 = 1`

Explanation:

The centre of the circle is the point on the positive part of the `y` axis which lies at distance `1` from the parabola.

Let `(a, a^2)` be any point on the parabola with `a != 0`

Then the tangent at `(a, a^2)` has slope `2a`.

So the normal at `(a, a^2)` has slope `-1/(2a)`

The equation of the normal can be written in point slope form as:

`y - a^2 = -1/(2a)(x - a)`

Putting `x=0` we find the `y`-intercept:

`y = a^2+1/2`

The distance between `(a, a^2)` and `(0, a^2+1/2)` is:

`sqrt((0-a)^2+((a^2+1/2)-a^2)^2) = sqrt(a^2+1/4)`

If this intercept is the centre of the circle then this distance is `1`:

`1 = sqrt(a^2+1/4)`

Squaring both sides:

`1 = a^2+1/4`

So the centre of the circle is:

`(0, a^2+1/2) = (0, (a^2+1/4)+1/4) = (0, 5/4)`

The standard form of the equation of a circle is:

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

where `(h, k)` is the centre of the circle and `r` the radius.

So for our unit circle, we have:

`(x-0)^2+(y-5/4)^2 = 1^2`

or more simply:

`x^2+(y-5/4)^2 = 1`

graph{(y-x^2)(x^2+(y-5/4)^2-1)(x^2+(y-5/4)^2-0.001)((abs(x)-sqrt(3)/2)^2+(y-3/4)^2-0.001)((y-3/4)+1/sqrt(3)(abs(x)-sqrt(3)/2))((y-3/4)-sqrt(3)(abs(x)-sqrt(3)/2))= 0 [-2.516, 2.484, -0.11, 2.39]}