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 __________.

Tangent of slope m to the parabola y^2=4 x is given by y=m x+1m and Tangent of slope m to the circle (x-4)^2+y^2=16 is given by y=m(x-4) 4 1+m^2 For common tangent aligned & 1m=-4 m 4 1+m^2 \\\\ & (1m+4 m)^2=( 4 1+m^2)^2 aligned On squaring both sides, we get aligned & =1m^2+1

Let a common tangent to the curves y2 4x and x 42 y2 16 touc: null (Mathematics)

Tangent of slope

m

to the parabola

y^2=4 x

is given by

y=m x+\frac{1}{m}

and

Tangent of slope

m

to the circle

(x-4)^2+y^2=16

is given by

y=m(x-4) \pm 4 \sqrt{1+m^2}

For common tangent

\begin{aligned} & \frac{1}{m}=-4 m \pm 4 \sqrt{1+m^2} \\\\ & \Rightarrow \left(\frac{1}{m}+4 m\right)^2=\left( \pm 4 \sqrt{1+m^2}\right)^2 \end{aligned}

On squaring both sides, we get

\begin{aligned} & =\frac{1}{m^2}+16 m^2+8=16+16 m^2 \\\\ & \Rightarrow \frac{1}{m^2} =8 \Rightarrow m= \pm \frac{1}{2 \sqrt{2}} \end{aligned}

Then, the point of contact on parabola is

(8,4 \sqrt{2})

Length of tangent

P Q

from

(8,4 \sqrt{2})

on the circle is

\begin{array}{ll} &\Rightarrow P Q=\sqrt{(8-4)^2+(4 \sqrt{2})^2-16} \\\\ &\Rightarrow P Q=\sqrt{16+32-16} \\\\ &\Rightarrow P Q=\sqrt{32} \Rightarrow(P Q)^2=32 \end{array}

Explanation

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