Question

The equation of the common tangent touching the circle ${{\left( x-3 \right)}^{2}}+{{y}^{2}}=9$ and parabola ${{y}^{2}}=4x$ above the x-axis is?
A. $\sqrt{3y}=3x+1$
B. $\sqrt{3y}=-\left( x+3 \right)$
C. $\sqrt{3y}=\left( x+3 \right)$
D. $\sqrt{3y}=-\left( 3x+1 \right)$

Answer 1

Hint: Compare the given equation of circle with standard equation of circle i.e.

${{\left( x-{{x}_{1}} \right)}^{2}}+{{\left( y-{{y}_{1}} \right)}^{2}}={{r}^{2}}$

Where $\left( {{x}_{1}},{{y}_{1}} \right)$ is centre and r is the radius. Equation of tangent for parabola ${{y}^{2}}=4ax$ is given as

$y=mx+\dfrac{a}{m}$

Any tangent to a circle will be perpendicular to the radius to that point and that radius will be the perpendicular distance from the centre to the tangent at that point.

Complete step-by-step answer:

Here, we have equation of circle and parabola as

$\begin{aligned}

& {{\left( x-3 \right)}^{2}}+{{y}^{2}}=9........................\left( i \right) \\

& {{y}^{2}}=4x............................\left( ii \right) \\

\end{aligned}$

Now, we know the standard equation of circle is given as

${{\left( x-{{x}_{1}} \right)}^{2}}+{{\left( y-{{y}_{_{1}}} \right)}^{2}}={{r}^{2}}..........................\left( iii \right)$

Where $\left( {{x}_{1}},{{y}_{1}} \right)$ is centre and r is the radius. So, we get centre of the circle given in the problem by comparing (i) and (ii) as

${{\left( x-3 \right)}^{2}}+{{\left( y-0 \right)}^{2}}={{\left( 3 \right)}^{2}}$

Centre = (3, 0), Radius = 3

And we can compare the equation of parabola ${{y}^{2}}=4x$ from the problem to the standard equation of parabola i.e. ${{y}^{2}}=4ax$ , which has vertex at (0, 0) and is symmetric about positive x-axis. So, we can get diagram as

Now, we know equation of tangent of parabola ${{y}^{2}}=4ax$ can be given as

$y=mx+\dfrac{a}{m}.......................\left( iv \right)$

Where m is the slope of tangent. Now, it is given that we need to determine the common tangent to the parabola and circle both. So, the tangent of equation (iv) will be a tangent for the circle as well.

Now, we know tangent to any point of circle is perpendicular to the radius to that point. So, the perpendicular distance from the centre of the circle to the tangent is the radius to the point tangency. So, we know perpendicular length of line Ax + By + C = 0 to the coordinate $\left( {{x}_{1}},{{y}_{1}} \right)$ can be given as

$\text{Perpendicular distance=}\left| \dfrac{A{{x}_{1}}+B{{y}_{1}}+C}{\sqrt{{{A}^{2}}+{{B}^{2}}}} \right|$ ………………………(v)

Hence, tangent for parabola ${{y}^{2}}=4x$ from equation (iv) can be given as

$y=mx+\dfrac{1}{m}...................\left( vii \right)$

Hence, the line represented in equation (viii) has a perpendicular length of ‘3’ from centre (3, 0). So, we can get from equation (v) as

$\text{Tangent}=mx-y+\dfrac{1}{m}=0$

Centre = (3, o)

Radius = Perpendicular length = 3. So, we get

$\begin{aligned}

& \left| \dfrac{m\left( 3 \right)-0+\dfrac{1}{m}}{\sqrt{1+{{m}^{2}}}} \right|=3 \\

& \left| \dfrac{3m+\dfrac{1}{m}}{\sqrt{1+{{m}^{2}}}} \right|=3 \\

\end{aligned}$

Squaring both the sides, we get

$\dfrac{{{\left( 3m+\dfrac{1}{m} \right)}^{2}}}{1+{{m}^{2}}}=9$

On cross multiplying we get

${{\left( 3m+\dfrac{1}{m} \right)}^{2}}=9\left( 1+{{m}^{2}} \right)$

We know

${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$

So we get

$\begin{aligned}

& {{\left( 3m \right)}^{2}}+{{\left( \dfrac{1}{m} \right)}^{2}}+2\times 3m\times \dfrac{1}{m}=9+9{{m}^{2}} \\

& 9{{m}^{2}}+\dfrac{1}{{{m}^{2}}}+6=9+9{{m}^{2}} \\

& \dfrac{1}{{{m}^{2}}}=3 \\

& {{m}^{2}}=\dfrac{1}{3} \\

& m=\pm \dfrac{1}{\sqrt{3}} \\

\end{aligned}$

Hence, we can get two common tangents from equation (vii) as

$y=\dfrac{x}{\sqrt{3}}+\sqrt{3},y=\dfrac{-1}{\sqrt{3}}x-\sqrt{3}$

Now, we can observe from the diagram there will be two common tangents for circle and parabola, one is above the x-axis with positive slope and another will be below the x-axis which will have negative slope. So, the equation of tangent with positive slope will be the answer to the problem. Hence,

$y=\dfrac{x}{\sqrt{3}}+\sqrt{3}$

Has positive slope (in comparison with equation of line y = mx + c). so, we get equation of tangent above x-axis as

$\begin{aligned}

& y=\dfrac{x}{\sqrt{3}}+\sqrt{3} \\

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

\end{aligned}$

So, option (c) is correct.

Note: We can suppose equation of tangent for circle as well as

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)\pm a\sqrt{{{m}^{2}}+1}$

Where $\left( {{x}_{1}},{{y}_{1}} \right)$ is centre. Now, satisfy this tangent as a tangent for parabola as well, where we need to substitute value of ‘x’ or ‘y’ from above equation to equation of parabola and equate the discriminant of formed quadrant to 0 because tangent and parabola will have only intersection point i.e. only one root of formed quadratic.

We can prove the equation $y=mx+\dfrac{a}{m}$ is tangent for ${{y}^{2}}=4ax$ as;

Suppose y = mx + c is tangent for ${{y}^{2}}=4ax$ . Put the value of y as mx + c to the equation of parabola i.e. ${{y}^{2}}=4ax$ . And now make the discriminant $\left( {{b}^{2}}-4ac \right)$ of the formed quadratic to 0 as quadratic should have equal roots. So, we get the value of ‘c’ as $\dfrac{a}{m}$ .