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Find the equation of the straight lines touching both ${{x}^{2}}+{{y}^{2}}={{a}^{2}}$ and ${{y}^{2}}=4ax$ .

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Hint: Find the centre, radius and slope of the circle and substitute in the equation of tangent with slope.

A tangent line is a line which locally touches a curve at one and only one point
We know that the equation of tangent$\Rightarrow y=mx+c..................\left( 1 \right)$
Which is also the slope intercept formula for a line $\Rightarrow y=mx+c$
Where $m$ is the slope of the line
$c$ is the y-intercept of the line
We have been given the general form of a circle
${{x}^{2}}+{{y}^{2}}={{a}^{2}}........................\left( 2 \right)$
Here the centre of the circle is $\left( 0,0 \right)$ and
Radius $=\sqrt{{{a}^{2}}}=a$
We know the equation of parabola $\Rightarrow {{y}^{2}}=4ax....................\left( 3 \right)$
The $y-$ intercept, $c=\dfrac{a}{m}$
Where $a$ is the radius of the circle and $m$ , the slope
$\therefore $ Equation $\left( 1 \right)$ can be written as
$y=mx+c$
$\Rightarrow \left( y=mx+\dfrac{a}{m} \right).................\left( 4 \right)$
The equation of tangent to circle with slope $m$ is given by the formula
$y=mx\pm a\sqrt{1+{{m}^{2}}}.......................\left( 5 \right)$
Now comparing both equation $\left( 4 \right)\And \left( 5 \right)$ and cancelling out like terms
$\begin{align}
  & mx+\dfrac{a}{m}=mx\pm a\sqrt{1+{{m}^{2}}} \\
 & \dfrac{a}{m}=\pm a\sqrt{1+{{m}^{2}}}\Rightarrow \dfrac{1}{m}=\pm \sqrt{1+{{m}^{2}}} \\
\end{align}$
Now squaring on both sides, we get
$\begin{align}
  & {{\left( \dfrac{1}{m} \right)}^{2}}={{\left( \pm \sqrt{1+{{m}^{2}}} \right)}^{2}} \\
 & \dfrac{1}{{{m}^{2}}}=1+{{m}^{2}} \\
 & \\
\end{align}$
Cross multiplying the above we get
$\begin{align}
  & {{m}^{2}}\left( 1+{{m}^{2}} \right)=1 \\
 & \Rightarrow {{m}^{4}}+{{m}^{2}}-1=0 \\
\end{align}$
$\therefore $ We get $\left( {{m}^{2}}+1 \right)\left( {{m}^{2}}-1 \right)=0$
We can remove the term $\left( {{m}^{2}}+1 \right)$
$\therefore \left( {{m}^{2}}-1 \right)=0\Rightarrow m=\pm 1$
Now substitute the value of $m$ in equation $\left( 4 \right)$
$\begin{align}
  & y=mx+\dfrac{a}{m} \\
 & y=\pm x+a \\
\end{align}$
Substitute the value of $m$ in equation$\left( 5 \right)$
\[\begin{align}
  & y=mx\pm a\sqrt{1+{{m}^{2}}} \\
 & y=\pm x\pm \sqrt{2a} \\
\end{align}\]
i.e. we got the required equation tangent.

Note: It’s important to remember the $2$ equation of tangent with which this question was solved.
$\begin{align}
  & y=mx+c\text{ or }y=mx+\dfrac{a}{m} \\
 & y=mx\pm a\sqrt{1+{{m}^{2}}} \\
\end{align}$
Misplacing this equation will generate wrong answers and the desired equation of tangent won’t happen.
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