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!!!
Best Answer
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.