Let r be the radius of the circle which touches the x-axis at the point (a,0), a

Question

Let $r$ be the radius of the circle which touches the $x$-axis at the point $(a,0)$, $a<0$, and the parabola $y^2=9x$ at the point $(4,6)$. Then $r$ is equal to \_\_\_\_\_\_\_\_.

Answer: 30

Accepted Answer

Circle: $(x-a)^2+(y-r)^2=r^2$ (centre $(a,r)$, tangent to $x$-axis). Through $(4,6)$:

$$a^2-8a-12r+52=0 \quad(1).$$

Tangent to the parabola at $(4,6)$: $6y=\dfrac92(x+4)\Rightarrow 3x-4y+12=0$, which is also tangent to the circle:

$$\frac{|3a-4r+12|}{5}=r \;\Rightarrow\; 3a+12=4r\pm5r.$$

Case $3a+12=9r$: gives $a=6$ (rejected on checking (1)). Case $3a+12=-r\Rightarrow a=\dfrac{-r-12}{3}$:

$$\left(\frac{-r-12}{3}\right)^2-8\left(\frac{-r-12}{3}\right)-12r+52=0 \;\Rightarrow\; r^2-60r+900=0 \;\Rightarrow\; (r-30)^2=0.$$

$$r=30 \quad(\text{with } a=-14).$$

Solutions · 1