How do you find the equation of the circle with a radius of 11 inscribed in the parabola y=x^2y=x2?

1 Answer
George C.
Mar 23, 2018

x^2+(y-5/4)^2 = 1x2+(y−54)2=1

Explanation:

The centre of the circle is the point on the positive part of the yy axis which lies at distance 11 from the parabola.

Let (a, a^2)(a,a2) be any point on the parabola with a != 0a≠0

Then the tangent at (a, a^2)(a,a2) has slope 2a2a.

So the normal at (a, a^2)(a,a2) has slope -1/(2a)−12a

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

y - a^2 = -1/(2a)(x - a)y−a2=−12a(x−a)

Putting x=0x=0 we find the yy-intercept:

y = a^2+1/2y=a2+12

The distance between (a, a^2)(a,a2) and (0, a^2+1/2)(0,a2+12) is:

sqrt((0-a)^2+((a^2+1/2)-a^2)^2) = sqrt(a^2+1/4)√(0−a)2+((a2+12)−a2)2=√a2+14

If this intercept is the centre of the circle then this distance is 11:

1 = sqrt(a^2+1/4)1=√a2+14

Squaring both sides:

1 = a^2+1/41=a2+14

So the centre of the circle is:

(0, a^2+1/2) = (0, (a^2+1/4)+1/4) = (0, 5/4)(0,a2+12)=(0,(a2+14)+14)=(0,54)

The standard form of the equation of a circle is:

(x-h)^2+(y-h)^2 = r^2(x−h)2+(y−h)2=r2

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

So for our unit circle, we have:

(x-0)^2+(y-5/4)^2 = 1^2(x−0)2+(y−54)2=12

or more simply:

x^2+(y-5/4)^2 = 1x2+(y−54)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]}

Related questions
See all questions in Standard Form of the Equation
Impact of this question
7083 views around the world
You can reuse this answer
Creative Commons License