Question

Find the equation of the common tangent touching of the circle $(x-3{)}^2+y^2=9$ and the parabola $y^2=4x$ above the $x$ -axis.

Answer 1

Given the equation------

$\begin{array}{c}Theparabola:y^2=4x\\ y=mx+\frac{1}{m}\\ y-mx-\frac{1}{m}=0|centerofcircle(3,0)\\ \Rightarrow \left|\frac{0-3m-\frac{1}{m}}{\sqrt{1+m^2}}\right|=3\\ bothsidesquaring....\\ \Rightarrow 9m^2+\frac{1}{m^2}+6=9(1+m^2)\\ \Rightarrow 9m^2+\frac{1}{m^2}+6=9+9m^{2}y=mx+\frac{1}{m}\\ \Rightarrow \frac{1}{m^2}=3\\ \therefore m=\pm \frac{1}{\sqrt{3}}\\ Now,wehavetwotangent--\\ let{T}_1:y=\frac{1}{\sqrt{3}}x+\sqrt{3}=\sqrt{3}y-x-3=0\\ T_2:y=-\frac{1}{\sqrt{3}}x-\sqrt{3}=\sqrt{3}y+x+3=0\\ lets,from\tan gent\sqrt{3}y-x-3=0touchabovethex-axis.\end{array}$