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

Mar 23, 2018

${x}^{2} + {\left(y - \frac{5}{4}\right)}^{2} = 1$

#### 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 $\left(a , {a}^{2}\right)$ be any point on the parabola with $a \ne 0$

Then the tangent at $\left(a , {a}^{2}\right)$ has slope $2 a$.

So the normal at $\left(a , {a}^{2}\right)$ has slope $- \frac{1}{2 a}$

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

$y - {a}^{2} = - \frac{1}{2 a} \left(x - a\right)$

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

$y = {a}^{2} + \frac{1}{2}$

The distance between $\left(a , {a}^{2}\right)$ and $\left(0 , {a}^{2} + \frac{1}{2}\right)$ is:

$\sqrt{{\left(0 - a\right)}^{2} + {\left(\left({a}^{2} + \frac{1}{2}\right) - {a}^{2}\right)}^{2}} = \sqrt{{a}^{2} + \frac{1}{4}}$

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

$1 = \sqrt{{a}^{2} + \frac{1}{4}}$

Squaring both sides:

$1 = {a}^{2} + \frac{1}{4}$

So the centre of the circle is:

$\left(0 , {a}^{2} + \frac{1}{2}\right) = \left(0 , \left({a}^{2} + \frac{1}{4}\right) + \frac{1}{4}\right) = \left(0 , \frac{5}{4}\right)$

The standard form of the equation of a circle is:

${\left(x - h\right)}^{2} + {\left(y - h\right)}^{2} = {r}^{2}$

where $\left(h , k\right)$ is the centre of the circle and $r$ the radius.

So for our unit circle, we have:

${\left(x - 0\right)}^{2} + {\left(y - \frac{5}{4}\right)}^{2} = {1}^{2}$

or more simply:

${x}^{2} + {\left(y - \frac{5}{4}\right)}^{2} = 1$

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]}