equation of common tangent touching circle and the parabola 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!!!
 A: 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.
A: 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!
A: \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}
