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  • The equation of the common tangent touching the circle   

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

Option: 1

\begin{aligned} & \sqrt{3} y=3 x+1 \\ \end{aligned}


Option: 2

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


Option: 3

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


Option: 4

\sqrt{3} y=-(3 x+1)


Answers (1)

best_answer

Any tangent to y^2=4 x is of the form y=m x+\frac{1}{m}  ( ) a = 1 and this touches the circle

\begin{aligned} &(x-3)^2+y^2=9\\ &\text { If, }_{,}\left|\frac{m(3)+\frac{1}{m}-0}{\sqrt{m^2+1}}\right|=3 \end{aligned}

[ center of the circle is (3,0) and radius is 3].

\begin{aligned} & \Rightarrow \frac{3 m^2+1}{m}= \pm 3 \sqrt{m^2+1} \\ \Rightarrow & 3 m^2+1= \pm 3 m \sqrt{m^2+1} \\ \Rightarrow & 9 m^4+1+6 m^2=9 m^2\left(m^2+1\right) \\ \Rightarrow & 9 m^4+1+6 m^2=9 m^4+9 m^2 \\ \Rightarrow & 3 m^2=1 \\ \Rightarrow & m= \pm \frac{1}{\sqrt{3}} \end{aligned}

If the tangent touches the parabola and circle above the X-axis, then slope m should be positive.

\therefore m=\frac{1}{\sqrt{3}}     and the equation is   y=\frac{1}{\sqrt{3}} x+\sqrt{3}

\Rightarrow \sqrt{3} y=x+3

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