Question

What is the largest value of `a` for which the circle `x^2+y^2=a^2` lies completely inside the parabola `y^2=4(x+4)` ?

Answer 1

`a=2\sqrt{3}`.

Explanation:

Let's work this out. We have

`y^2=a^2-x^2`

`y^2=4(x+4)`

So then when the circle intersects the parabola the values of `y^2` must agree:

`a^2-x^2=4(x+4)`

`x^2+4x+(16-a^2)=0`

We now find the positive value of `a` where the quadratic equation has just one root, where the circle will just touch the parabola. To get that the discriminant must be zero:

(Middle coefficient)^2-(4×(product of other coefficients))=0

`4^2-(4×(16-a^2))=0`

`a^2=12=2^2×3`

`a=2\sqrt{3}`.