Question

The equations of the common tangent touching the circle \[{{(x-3)}^{2}}+{{y}^{2}}=9\] and parabola \[{{y}^{2}}=4x\] above the x-axis is
(a) \[\sqrt{3}y=3x+1\]
(b) \[\sqrt{3}y=-(x+3)\]
(c) \[\sqrt{3}y=x+3\]
(d) \[\sqrt{3}y=-(3x+1)\]

Answer 1

Hint: To find the equation of the common tangent to both the curves, write the equation of the tangent to both the curves in slope form, and then compare the slope and intercept of both equations. Simplify the equations to calculate the slope of the tangent. Substitute the value of the slope of the tangent to find the equation of common tangent.

Complete step by step answer:
We have a circle \[{{(x-3)}^{2}}+{{y}^{2}}=9\] and a parabola \[{{y}^{2}}=4x\]. We need to find the equation of common tangents to both the curves.
 We know that the equation of the tangent to the parabola of the form \[{{y}^{2}}=4ax\] with slope m is \[y=mx+\dfrac{a}{m}\].
Substituting the value \[a=1\] in the above equation, we get \[y=mx+\dfrac{1}{m}.....\left( 1 \right)\] as the equation of the tangent to the parabola \[{{y}^{2}}=4x\].
We know that the equation of the tangent to the circle of the form \[{{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}\] with slope m is \[y-k=m(x-h)\pm r\sqrt{1+{{m}^{2}}}\].
Substituting the values \[h=3,k=0,r=3\], we get \[y=m(x-3)\pm 3\sqrt{1+{{m}^{2}}}.....\left( 2 \right)\] as the equation of the tangent to the circle \[{{(x-3)}^{2}}+{{y}^{2}}={{3}^{2}}\].
We know that the equation \[(1)\] and \[(2)\] represent the same line.
Comparing the intercept of both the lines, we get \[\pm 3\sqrt{1+{{m}^{2}}}-3m=\dfrac{1}{m}\].
\[\Rightarrow \pm 3\sqrt{1+{{m}^{2}}}=3m+\dfrac{1}{m}\]
Squaring on both sides we get \[9(1+{{m}^{2}})=9{{m}^{2}}+\dfrac{1}{{{m}^{2}}}+6\].
\[\Rightarrow 9+9{{m}^{2}}=9{{m}^{2}}+\dfrac{1}{{{m}^{2}}}+6\]
Solving the above equation, we get \[3=\dfrac{1}{{{m}^{2}}}\].
Taking square root on both sides, we have $m=\pm \dfrac{1}{\sqrt{3}}$.
So, the possible equations of tangents to the curves are \[y=\dfrac{1}{\sqrt{3}}x+\sqrt{3}\] and \[y=-\dfrac{1}{\sqrt{3}}x-\sqrt{3}\].
As the circle and parabola lie in the right half-plane, the tangent with the positive slope intersects the curves in the first quadrant. While the tangent with negative slope intersects the curves in the fourth quadrant.

We know that our tangent lies above the x-axis.
So, the equation of tangent intersecting our curves is \[\sqrt{3}y=x+3\]. We will reject the other equation of tangent.
Hence, the correct answer is \[\sqrt{3}y=x+3\]

So, the correct answer is “Option C”.

Note: You won’t get a correct answer if you don’t consider the fact that tangent lies above the x-axis. We can also solve this question by finding the point of intersection of circle and parabola and then writing the equation of tangent in point slope form.