Equation of common tangent.

Question

What is the equation of common tangent to the circle $(x-3)^2+y^2=9$ and parabola $y^2=4x$.$$My Try$$ So equation of tangent at point $(x_1,y_1)$ is $xx_1+yy_1-3(x+x_1)=0,yy_1=2(x+x_1)$ for circle,parabola respectively.Then $x_1=\frac{x}{1-x}$ but now I am thinking how to proceed.

Answer 1

First of all, $x=0$ is a common tangent.

Let $(p,q)$ be the tanget point on the circle, and let $(s,t)$ be the tangent point on the parabola. Here, $p,q,s,t\not=0$.

The equation of the tangent line at $(p,q)$ is given by $$(p-3)(x-3)+qy=9\implies y=\frac{3-p}{q}x+\frac{3p}{q}$$ Also, the equation of the tangent line at $(s,t)$ is given by $$y-t=\frac{2}{t}(x-s)\implies y=\frac 2tx+\frac{2s}{t}$$

So, we have the following system : $$(p-3)^2+q^2=9\tag1$$ $$t^2=4s\tag2$$ $$\frac{3-p}{q}=\frac{2}{t}\tag3$$ $$\frac{3p}{q}=\frac{2s}{t}\tag4$$ From $(3)$, $$t=\frac{2q}{3-p}\tag5$$ From $(4)(5)$, $$s=\frac{3p}{3-p}\tag6$$ From $(1)(2)(5)(6)$, $$\left(\frac{2q}{3-p}\right)^2=4\cdot\frac{3p}{3-p}\implies p=\frac 32$$ since $p\not=0$.

Therefore, we have $$(p,q,s,t)=\left(\frac 32,\pm\frac{3\sqrt 3}{2},3,\pm 2\sqrt 3\right)$$ Thus, the equations of the common tangents are $$\color{red}{x=0,\qquad y=\pm\frac{1}{\sqrt 3}x\pm\sqrt 3}$$

Answer 2

hint : take tangent equation of parabola and find it's perpendicular distance from point $(3,0)$ and equate it to $3$ . https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_line