A line is a common tangent to the circle (x-3)^2+y^2=9 and the parabola y^2=4x.

Question

A line is a common tangent to the circle $(x-3)^2+y^2=9$ and the parabola $y^2=4x$. If the two points of contact $(a,b)$ and $(c,d)$ are distinct and lie in the first quadrant, then $2(a+c)$ is equal to \_\_\_\_\_\_\_\_.

Answer: 9

Accepted Answer

Tangent to $y^2=4x$: $y=mx+\dfrac1m$. Distance from $(3,0)$ to $m^2x-my+1=0$ equals $3$:
$$\frac{|3m^2+1|}{\sqrt{m^2+m^4}}=3\ \Longrightarrow\ (3m^2+1)^2=9(m^2+m^4)\ \Longrightarrow\ m=\frac1{\sqrt3}.$$

Point of contact on parabola: $(c,d)=\left(\dfrac1{m^2},\dfrac2m\right)=(3,2\sqrt3)$.

Substituting the tangent $y=\frac1{\sqrt3}x+\sqrt3$ into the circle: $\dfrac43x^2-4x+3=0$, a repeated root at $x=\dfrac32=a$.

$$2(a+c)=2\left(\frac32+3\right)=9.$$

Solutions · 1