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

1 Answer
Mar 23, 2018

Answer:

#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]}