Let a common tangent to the curves y^2=4x and | vedclass

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(D) The general equation of a tangent with slope $m$ to the circle $(x-4)^2+y^2=16$ is $y=m(x-4) \pm 4\sqrt{1+m^2}$.The general equation of a tangent

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$32$

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(D) The general equation of a tangent with slope $m$ to the circle $(x-4)^2+y^2=16$ is $y=m(x-4) \pm 4\sqrt{1+m^2}$.The general equation of a tangent with slope $m$ to the parabola $y^2=4x$ is $y=mx+\frac{1}{m}$.For a common tangent,the constant terms must be equal: $\frac{1}{m} = -4m \pm 4\sqrt{1+m^2}$.Rearranging gives $\frac{1}{m} + 4m = \pm 4\sqrt{1+m^2}$. Squaring both sides: $(\frac{1}{m} + 4m)^2 = 16(1+m^2) \implies \frac{1}{m^2} + 16m^2 + 8 = 16 + 16m^2$.This simplifies to $\frac{1}{m^2} = 8$,so $m^2 = \frac{1}{8}$,which means $m = \pm \frac{1}{2\sqrt{2}}$.The point of contact $P$ on the parabola $y^2=4x$ is $(\frac{1}{m^2}, \frac{2}{m}) = (8, \pm 4\sqrt{2})$.The length of the tangent segment $PQ$ between the point of contact $P$ on the parabola and the point of contact $Q$ on the circle is given by the length of the tangent from $P$ to the circle.For a point $(x_1, y_1)$ and circle $(x-4)^2+y^2-16=0$,the length of the tangent is $\sqrt{(x_1-4)^2 + y_1^2 - 16}$.Substituting $P(8, 4\sqrt{2})$: $PQ = \sqrt{(8-4)^2 + (4\sqrt{2})^2 - 16} = \sqrt{16 + 32 - 16} = \sqrt{32}$.Therefore,$(PQ)^2 = 32$.

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 $..........$.

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