If it is given parabola:
$${y}^2 = 4x$$
How can I find a equation circle (center on x axis) that thouch parabola from inside? $$r=2(sqrt){5}$$
I have done next:
$$ y^2 = 2px $$ $$ y^2=2*2*x$$ $$ p=2$$ $$ r^2=(x-p)^2+(y-q)^2$$ $$ 20=x^2-4x+4+y^2$$
What have I done wrong?
On
Parabola: $y^2=4x$, circle: $20=(x-a)^2+y^2$.
We take implicit derivatives:
$$2yy'=4$$
$$0=2(x-a)+2yy'$$
In a touching point the derivatives (i.e. slopes of tangent line) are equal, so $$2(x-a)+4=0$$ $$x=a-2$$
And we plug this back into $\begin{cases}y^2=4x\\20=(x-a)^2+y^2\end{cases}$
beacuse the given curves intersect.
$$\begin{cases}y^2=4a-8\\20=(-2)^2+4a-8\end{cases}$$
$$4a=24$$
$$a=6$$
Let us assume the center of the circle at $(p,0)$. The intersections between the circle and the parabola are solutions of
$$\begin{cases}y^2=4x,\\(x-p)^2+y^2=20.\end{cases}$$
We eliminate $p$ and get a quadratic equation in $x$,
$$(x-p)^2+4x-20=0.$$
Now it suffices to express that the root is double, by canceling the discriminant:
$$6-p=0.$$