Animated Solution for Mathematics - Conic Sections: Let a common tangent to the curves $y^2=4x$ and $(x-4{)}^2+y^2=16$ touch the curves at the points $P$ and $Q$. Then $(PQ{)}^2$ is equal to ________.

Welcome, future engineer. Today, we are not just solving a coordinate geometry problem; we are choreographing a dance between two fundamental shapes: the parabola and the circle.

When you look at the equations $y^2=4x$ and $(x-4{)}^2+y^2=16$, do not just see variables and constants. See a parabola, opening its arms to the right, and a circle, resting comfortably with its center at $(4,0)$ and a radius of $4$.

Our mission is to find the common tangent that touches both, and then calculate the squared distance between the points of contact, $P$ and $Q$. This is a classic JEE Advanced problem that tests your ability to bridge algebra with geometric intuition.

To connect these two curves, we need a common thread—a line that satisfies the tangency conditions for both. We start with the parabola $y^2=4x$.

In the world of conic sections, the tangent to a parabola $y^2=4ax$ with a slope $m$ is given by the elegant equation $y=mx+\frac{a}{m}$. Here, our $a$ is $1$. Thus, our tangent line is $y=mx+\frac{1}{m}$.

To make this line easier to work with, let us rearrange it into the general form $Ax+By+C=0$. Multiplying by $m$, we get $my=m^{2}x+1$, or $m^{2}x-my+1=0$. This line is our protagonist.

Now, we turn our attention to the circle $(x-4{)}^2+y^2=16$. For our line $m^{2}x-my+1=0$ to be a tangent to this circle, it must maintain a very specific relationship with the circle's center.

Geometry tells us that the perpendicular distance from the center of the circle to the tangent line must be exactly equal to the radius of the circle. The center of our circle is $(4,0)$ and the radius $R$ is $4$.

Using the perpendicular distance formula, we substitute our values:

This simplifies to:

We square both sides to clear the radical:

Expanding the left side, we get $16m^4+8m^2+1=16m^4+16m^2$. Notice how the $16m^4$ terms cancel out perfectly. We are left with $8m^2=1$, which means $m^2=\frac{1}{8}$.

With $m^2=\frac{1}{8}$, we have unlocked the key to the coordinates. For the parabola, the point of contact $P$ is given by $(\frac{a}{m^2},\frac{2a}{m})$.

Substituting $a=1$ and $m^2=\frac{1}{8}$, the $x$-coordinate is $x_P=\frac{1}{1/8}=8$. For the $y$-coordinate, taking the positive case for $m=\frac{1}{2\sqrt{2}}$, we get $y_P=\frac{2}{1/(2\sqrt{2})}=4\sqrt{2}$. So, $P=(8,4\sqrt{2})$.

Now for point $Q$ on the circle. $Q$ is the foot of the perpendicular from the center $(4,0)$ to the line $m^{2}x-my+1=0$. With $m^2=\frac{1}{8}$ and $m=\frac{1}{2\sqrt{2}}$, the line becomes $x-2\sqrt{2}y+8=0$.

Using the foot of the perpendicular formula, we find:

We have reached the final stretch. We need $(PQ{)}^2$. Using the distance formula, $(PQ{)}^2=(x_P-x_Q{)}^2+(y_P-y_Q{)}^2$.

Substituting our coordinates:

This simplifies to:

Dividing $288$ by $9$ gives us exactly 32.

There it is. The complexity of the curves, the algebra of the tangents, and the geometry of the circle all collapse into a single, elegant integer: 32.