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

##### 1 Answer

#### Explanation:

The centre of the circle is the point on the positive part of the

Let

Then the tangent at

So the normal at

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

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

Putting

#y = a^2+1/2#

The distance between

#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 = 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

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]}