Animated Solution for Mathematics - Conic Sections: Let $r$ be the radius of the circle, which touches $x$-axis at point $(a,0)$, $a<0$ and the parabola $y^2=9x$ at the point $(4,6)$. Then $r$ is equal to ________

Imagine you are standing on a coordinate plane. Before you lies the elegant curve of a parabola, defined by the equation $y^2=9x$. It is a smooth, sweeping arc that opens to the right.

Now, imagine a circle, perfect and symmetrical, resting gently against this parabola at the specific point $P(4,6)$. This circle is anchored, touching the $x$-axis at a point $(a,0)$ where $a<0$.

To solve this, we must bridge the gap between calculus and geometry.

When two curves kiss at a single point, they share a common tangent. This is the foundation of our journey. If they share a tangent, they must also share a common normal.

This normal line is the "spine" of our circle—it is the line that passes through the center of the circle and is perpendicular to the tangent at the point of contact. To find this normal, we first need the slope of the tangent.

We take our parabola $y^2=9x$ and differentiate it with respect to $x$. Using the chain rule, we get:

At our point of contact $P(4,6)$, the slope of the tangent $m_t$ is:

The normal is perpendicular to this tangent. The rule of perpendicular slopes tells us that the product of the slopes must be $-1$. Thus, the slope of our normal $m_n$ is the negative reciprocal of $\frac{3}{4}$, which is $-\frac{4}{3}$.

With a point $(4,6)$ and a slope of $-\frac{4}{3}$, we write the equation of the normal using the point-slope form:

Multiplying by $3$ and rearranging, we arrive at the linear equation:

The circle touches the $x$-axis at $(a,0)$. Geometrically, this means the center of the circle must lie directly above this point of contact. If the circle has a radius $r$, then the center $C$ must be at $(a,r)$.

Because the normal line must pass through the center of the circle, the coordinates $(a,r)$ must satisfy the equation $4x+3y-34=0$. Substituting $x=a$ and $y=r$, we get:

We have successfully reduced our variables. Everything is now a function of the radius $r$.

The distance from the center $C(a,r)$ to the point of contact $P(4,6)$ must be exactly the radius $r$. Using the distance formula:

Substitute our expression for $a$ into this equation:

Simplifying the first term, we get:

The equation becomes:

Notice how the $r^2$ terms on both sides cancel out. We are left with:

Multiplying by $16$ to clear the fraction and expanding, we obtain the quadratic equation:

Solving the quadratic $3r^2-100r+300=0$ is straightforward. We factor it as:

This gives us two potential radii: $r=\frac{10}{3}$ and $r=30$. We must respect the constraint $a<0$.

If we test $r=\frac{10}{3}$, we find $a=\frac{34-3(10/3)}{4}=\frac{24}{4}=6$, which is positive. If we test $r=30$, we find:

Since $-14<0$, this value satisfies our condition. The radius of our circle is 30.