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

Answer

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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.

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.

Last updated date: 01st Oct 2023

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