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

Question

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

Answer: 30

Accepted Answer

$\begin{gathered} (x-a{)}^{2}+(y-r{)}^{2}={r}^{2} \ (4-a{)}^{2}+(6-r{)}^{2}={r}^{2} \ 16+{a}^{2}-8a+36+{r}^{2}-12r={r}^{2} \ {a}^{2}-8a-12r+52=0 \end{gathered}$

Tangent to parabola at $(4,6)$ is

$6.4=9.(\frac{x+4}{2})3x-4y+12=0$

This is also tangent to the circle

$\begin{gathered} 	herefore \, CP=r \ \frac{3a-4r+12}{5}=\pm r \end{gathered}$

$3a+12=4r\pm 5r \begin{gathered} ar \ -r \end{gathered}\,$

equation of circle is

$(x-a{)}^{2}+(y-r{)}^{2}={r}^{2}$

satsty $P(4,6)\Rightarrow {a}^{2}-8a-12r+52=0\,$

From equation (1)

If $a+4=3r$ then $a=+6$ (rejected)

If $3a+12=-r$ then $a=-14$ and $r=30$

Solutions · 1