1. Understanding the Curves and the Objective

We are given two curves:

1. A parabola: ${y^2} = 4x$
2. A circle: ${(x - 4)^2} + {y^2} = 16$

Our goal is to find a line that is tangent to both curves (a common tangent). Let this tangent touch the parabola at point P and the circle at point Q. Finally, we need to calculate the squared distance between P and Q, i.e., $(PQ)^2$.

2. General Equation of a Tangent to the Parabola

Key Concept: The equation of a tangent with slope $m$ to a parabola of the form $y^2 = 4ax$ is given by $y = mx + \frac{a}{m}$. The point of contact on the parabola for this tangent is $P\left(\frac{a}{m^2}, \frac{2a}{m}\right)$.

Step-by-step Derivation:

• Identify 'a' for the parabola: The given parabola is $y^2 = 4x$. Comparing this with the standard form $y^2 = 4ax$, we see that $4a = 4$, which implies $a = 1$.
• Write the general tangent equation: Substitute $a=1$ into the tangent formula: $y = mx + \frac{1}{m} \quad (*)$ This equation represents any non-vertical tangent line to the parabola $y^2=4x$. We need to find the specific value(s) of $m$ for which this line is also tangent to the given circle.
• Write the general point of contact P: For this tangent, the point of contact on the parabola is: $P\left(\frac{1}{m^2}, \frac{2}{m}\right)$

3. Condition for the Line to be Tangent to the Circle

Key Concept: A line $Lx + My + N = 0$ is tangent to a circle with center $(h, k)$ and radius $R$ if and only if the perpendicular distance from the center of the circle to the line is equal to the radius of the circle. The distance formula is $d = \frac{|Lh + Mk + N|}{\sqrt{L^2 + M^2}}$.

Step-by-step Derivation:

• Identify center and radius of the circle: The given circle is $(x - 4)^2 + y^2 = 16$.
  • Comparing this with the standard form $(x-h)^2 + (y-k)^2 = R^2$, we identify the center as $(h, k) = (4, 0)$ and the radius as $R = \sqrt{16} = 4$.
• Rewrite the tangent equation in general form: Our tangent equation from step 2 is $y = mx + \frac{1}{m}$.
  • Rearrange it to the form $Lx + My + N = 0$: $mx - y + \frac{1}{m} = 0$
  • Here, $L=m$, $M=-1$, $N=\frac{1}{m}$.
• Apply the tangency condition: The distance from the center $(4,0)$ to the line $mx - y + \frac{1}{m} = 0$ must be equal to the radius $R=4$. $\frac{|m(4) - (0) + \frac{1}{m}|}{\sqrt{m^2 + (-1)^2}} = 4$ $\frac{|4m + \frac{1}{m}|}{\sqrt{m^2 + 1}} = 4$
• Solve for m:
  • Square both sides to eliminate the absolute value and square root: $\left(4m + \frac{1}{m}\right)^2 = 16(m^2 + 1)$ $\left(\frac{4m^2 + 1}{m}\right)^2 = 16(m^2 + 1)$ $\frac{(4m^2 + 1)^2}{m^2} = 16(m^2 + 1)$ $(16m^4 + 8m^2 + 1) = 16m^2(m^2 + 1)$ $16m^4 + 8m^2 + 1 = 16m^4 + 16m^2$ This step is crucial: we are expanding both sides to simplify and solve for $m$. $8m^2 + 1 = 16m^2$ $1 = 8m^2$ $m^2 = \frac{1}{8}$ $m = \pm \frac{1}{\sqrt{8}} = \pm \frac{1}{2\sqrt{2}}$
  • We have found two possible slopes for the common tangents. This is expected, as a parabola and a circle typically have two common tangents (one with a positive slope and one with a negative slope, symmetric about the x-axis in this case due to the symmetry of the curves). We can choose either value of $m$ since $(PQ)^2$ will be the same for both symmetric tangents. Let's choose $m = \frac{1}{2\sqrt{2}}$.

4. Finding the Specific Common Tangent and Points of Contact P and Q

• Common Tangent Equation: Substitute $m = \frac{1}{2\sqrt{2}}$ into $y = mx + \frac{1}{m}$: $y = \frac{1}{2\sqrt{2}}x + \frac{1}{\frac{1}{2\sqrt{2}}}$ $y = \frac{1}{2\sqrt{2}}x + 2\sqrt{2}$ This is the equation of one of the common tangents.

• Point P (on Parabola): Substitute $m = \frac{1}{2\sqrt{2}}$ and $a=1$ into $P\left(\frac{a}{m^2}, \frac{2a}{m}\right)$: $P\left(\frac{1}{(1/(2\sqrt{2}))^2}, \frac{2}{1/(2\sqrt{2})}\right)$ $P\left(\frac{1}{1/8}, 2 \cdot 2\sqrt{2}\right)$ $P(8, 4\sqrt{2})$

• Point Q (on Circle): Method: The radius from the center of the circle to the point of contact Q is perpendicular to the tangent line.

  • The center of the circle is $C(4,0)$.
  • The slope of the tangent is $m = \frac{1}{2\sqrt{2}}$.
  • The slope of the radius $CQ$ must be $m_{CQ} = -\frac{1}{m} = -2\sqrt{2}$.
  • Equation of the line $CQ$: $y - 0 = -2\sqrt{2}(x - 4)$ $y = -2\sqrt{2}x + 8\sqrt{2} \quad (**)$
  • To find Q, we solve the system of equations formed by the tangent line $(*)$ and the radius line $(**)$: $y = \frac{1}{2\sqrt{2}}x + 2\sqrt{2}$ $y = -2\sqrt{2}x + 8\sqrt{2}$
  • Equate the expressions for $y$: $\frac{1}{2\sqrt{2}}x + 2\sqrt{2} = -2\sqrt{2}x + 8\sqrt{2}$ Our goal is to find the $x$-coordinate of Q. $\frac{1}{2\sqrt{2}}x + 2\sqrt{2}x = 8\sqrt{2} - 2\sqrt{2}$ $\left(\frac{1}{2\sqrt{2}} + 2\sqrt{2}\right)x = 6\sqrt{2}$ $\left(\frac{1 + (2\sqrt{2})^2}{2\sqrt{2}}\right)x = 6\sqrt{2}$ $\left(\frac{1 + 8}{2\sqrt{2}}\right)x = 6\sqrt{2}$ $\frac{9}{2\sqrt{2}}x = 6\sqrt{2}$ $x = \frac{6\sqrt{2} \cdot 2\sqrt{2}}{9} = \frac{6 \cdot 2 \cdot 2}{9} = \frac{24}{9} = \frac{8}{3}$
  • Now, substitute $x = \frac{8}{3}$ back into either tangent equation or radius equation to find $y$. Using the radius equation $y = -2\sqrt{2}x + 8\sqrt{2}$: $y = -2\sqrt{2}\left(\frac{8}{3}\right) + 8\sqrt{2}$ $y = -\frac{16\sqrt{2}}{3} + \frac{24\sqrt{2}}{3}$ $y = \frac{8\sqrt{2}}{3}$
  • So, the point of contact on the circle is $Q\left(\frac{8}{3}, \frac{8\sqrt{2}}{3}\right)$.

Common Mistake Alert: When finding the point of contact on the circle, simply substituting the tangent equation into the circle's equation and solving for $x$ (which would yield a quadratic with a single root) is also valid, but often more computationally intensive. The perpendicular radius method is usually more efficient.

5. Calculating $(PQ)^2$

Key Concept: The squared distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $D^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$.

Step-by-step Calculation:

• We have $P(8, 4\sqrt{2})$ and $Q\left(\frac{8}{3}, \frac{8\sqrt{2}}{3}\right)$.
• Calculate the difference in x-coordinates: $x_P - x_Q = 8 - \frac{8}{3} = \frac{24 - 8}{3} = \frac{16}{3}$
• Calculate the difference in y-coordinates: $y_P - y_Q = 4\sqrt{2} - \frac{8\sqrt{2}}{3} = \frac{12\sqrt{2} - 8\sqrt{2}}{3} = \frac{4\sqrt{2}}{3}$
• Now, calculate $(PQ)^2$: $(PQ)^2 = \left(\frac{16}{3}\right)^2 + \left(\frac{4\sqrt{2}}{3}\right)^2$ $(PQ)^2 = \frac{256}{9} + \frac{16 \cdot 2}{9}$ $(PQ)^2 = \frac{256}{9} + \frac{32}{9}$ $(PQ)^2 = \frac{256 + 32}{9}$ $(PQ)^2 = \frac{288}{9}$ $(PQ)^2 = 32$

6. Summary and Key Takeaway

The squared distance $(PQ)^2$ between the points of contact is 32.

This problem beautifully combines concepts from both parabolas and circles. The key steps involved:

1. Formulating the general tangent equation for the simpler curve (parabola).
2. Applying the tangency condition (distance from center to line equals radius) for the more complex curve (circle) to find the slope(s) of the common tangent(s).
3. Determining the specific points of contact on each curve using their respective properties (point of contact formula for parabola, perpendicular radius for circle).
4. Finally, using the distance formula to calculate the required squared distance.

Tip for JEE: Always try to visualize the curves. The parabola $y^2=4x$ opens to the right, and the circle $(x-4)^2+y^2=16$ has its center at $(4,0)$ and radius 4. Notice that the circle's leftmost point is $(0,0)$, which is the vertex of the parabola. This visual understanding helps anticipate the number of common tangents and check the reasonableness of your calculated points.