To find the square of the length of the line segment \(PQ\), we first need to analyze the given curves and determine their common tangent. The curves are \(y^2 = 4x\) and \((x-4)^2 + y^2 = 16\).
1. **Curve 1: Parabola** - Given by \(y^2 = 4x\). - Tangent line form at point \((x_1, y_1)\) is \(yy_1 = 2(x + x_1)\).
2. **Curve 2: Circle** - Given by \((x-4)^2 + y^2 = 16\). - Tangent line form at point \((x_2, y_2)\) is \((x_2-4)(x-4) + y_2y = 16\).
For these tangents to be common, their equations must be identical. Assume the line has slope \(m\) and intersect both curves. The tangent to the parabola can be written as \(y = mx + c\). - **For Parabola**: \(yy = 2(x + x_1) \Rightarrow y = mx + \frac{c}{m} = mx + \frac{y_1}{m}\)
- **For Circle**: The center of the circle is \((4,0)\) with radius \(4\). The perpendicular distance from center to the line is: \(\left|\frac{4m+c}{\sqrt{m^2+1}}\right| = 4\) Solving gives \(c^2 = 16(m^2+1)\).
Find common tangent as: tangent equations must equate. - **Solving**: \(m = 1\), \(c = 4\) satisfies both curves.
Equations are: \(y = x + 4\) is for both curves; find points of contact.
3. **Find Points P and Q**: - **For Parabola**: Using \(y = x + 4\) in \(y^2 = 4x\), substituting \(x+4\) for \(y\), gives: \((x+4)^2 = 4x\) \(x^2 + 8x + 16 = 4x\) Solving, \(x^2 + 4x + 16 = 0\) gives \(x = 4\), then \(y = 8\) at \(P(4,8)\).
- **For Circle**: Substituting \(y = x + 4\) in \((x-4)^2 + y^2 = 16\) and solve: \((x-4)^2 + (x+4)^2 = 16\) Expand to get \(2x^2 + 2x + 16 = 16\) x = -4 results in \(y=0\) at \(Q(-4, 0)\).
4. **Calculate \(PQ\) Distance**: Use the distance formula between points \(P(4,8)\) and \(Q(-4,0)\): \(d = \sqrt{(-4 - 4)^2 + (0 - 8)^2}\) \(d = \sqrt{64 + 64} = \sqrt{128}\) \(PQ = \sqrt{128} = 8\sqrt{2}\)
5. **Find \((PQ)^2\)**: \((PQ)^2 = (8\sqrt{2})^2 = 128\) which is clearly between 32 and 32.
