equation of common tangent touching circle and the parabola

Question

I am trying to find the equation of the common tangent touching the circle $(x-3)^2+y^2=9$ and the parabola $y^2=4x$ above the x-axis is :

Equation of tangent on parabola: $y=mx+a/m$ here a = 1; so $y = mx + 1/m$

centre of the circle (3,0) and r = 3. Now

distance of point from line is, $3 = |3m+1/m|/(1+m^2)^(1/2)$

I am having difficulty in finding the value of m. Please help me!!!

Answer 1

Instead of calculating the distance between the line and the center of the circle, and setting that equal to $3$ (which does work, mind you), I would find it more natural to take your expression for $y$ (i.e. $mx + 1/m$), insert it for $y$ in the equation for the circle, expand, and see for what values of $m$ the quadratic equation in $x$ that you get has exactly one solution. This way you're not relying on the second figure being a circle, only that it's defined by a second-degree expression.

Added details (assuming the calculations you have done are correct; I haven't checked them myself)
Making the substitution mentioned above, we get $$ (x-3)^2 + (mx + 1/m)^2 = 9\\ x^2 - 6x + 9 + m^2x^2 + 2x + 1/m^2 = 9\\ (1+m^2)x^2 - 4x + 1/m^2 = 0 $$ This quadratic equation has exactly one solution iff $4^2-4(1+m^2)/m^2 = 0$. This means $$ 4^2 = 4(1+m^2)/m^2\\ 4 = 1+1/m^2\\ \frac13 = m^2\\ m = \pm\frac1 {\sqrt3} $$ The two different solutions of $m$ correspond to the fact that the tangent could be either below or above the $x$-axis.

Answer 2

Hint:

Your work is correct. Now you have to solve the equation $$ \frac{|3m+\frac{1}{m}|}{\sqrt{1+m^2}}=3 \iff |3m+\frac{1}{m}|=3\sqrt{1+m^2} $$

Squaring the equation you have $$ \left(3m+\frac{1}{m}\right)^2=9(1+m^2) $$

easy to solve because the therm in $m^2$ vanish!

Answer 3

\begin{align} \frac{\displaystyle \left|3m+\frac{1}{m}\right|}{\sqrt{m^2+1}}&=3\\ \left(3m+\frac{1}{m}\right)^2&=9(m^2+1)\\ 9m^2+6+\frac{1}{m^2}&=9m^2+9\\ \frac{1}{m^2}&=3\\ m&=\frac{\pm1}{\sqrt{3}} \end{align}