What is the equation of the common tangent to | vedclass

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(C) Any tangent to the parabola $y^2 = 4x$ is of the form $y = mx + \frac{1}{m}$.It touches the circle $(x - 3)^2 + y^2 = 9$ (center $(3, 0)$,radius $

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$\sqrt{3}y = (x + 3)$

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(C) Any tangent to the parabola $y^2 = 4x$ is of the form $y = mx + \frac{1}{m}$.It touches the circle $(x - 3)^2 + y^2 = 9$ (center $(3, 0)$,radius $3$) if the perpendicular distance from the center to the line is equal to the radius:$3 = \left| \frac{m(3) - 0 + \frac{1}{m}}{\sqrt{m^2 + 1}} \right|$Squaring both sides:$9(m^2 + 1) = (3m + \frac{1}{m})^2$$9m^2 + 9 = 9m^2 + 6 + \frac{1}{m^2}$$3 = \frac{1}{m^2}$ $\Rightarrow m^2 = \frac{1}{3}$ $\Rightarrow m = \pm \frac{1}{\sqrt{3}}$For the tangent above the $X$-axis,we take the positive slope $m = \frac{1}{\sqrt{3}}$:$y = \frac{1}{\sqrt{3}}x + \frac{1}{1/\sqrt{3}}$$y = \frac{1}{\sqrt{3}}x + \sqrt{3}$Multiplying by $\sqrt{3}$:$\sqrt{3}y = x + 3$

What is the equation of the common tangent to the circle $(x - 3)^2 + y^2 = 9$ and the parabola $y^2 = 4x$ above the $X$-axis?

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