Let AB be a chord of length 12 of the circle (x-2)^2+(y+1)^2=1694.

Question

Let $AB$ be a chord of length $12$ of the circle $(x-2)^2+(y+1)^2=\dfrac{169}4$. If tangents drawn to the circle at points $A$ and $B$ intersect at the point $P$, then five times the distance of point $P$ from chord $AB$ is equal to \_\_\_\_\_\_\_\_.

Answer: 72

Accepted Answer

$r=\frac{13}2$. Distance from centre to chord: $d=\sqrt{r^2-6^2}=\sqrt{\frac{169}4-36}=\sqrt{\frac{25}4}=\frac52$.

Using $OP\cdot OM=r^2$ (pole-polar relation), $OP=\dfrac{r^2}d$, so distance from $P$ to chord:
$$OP-d=\frac{r^2-d^2}d=\frac{36}{5/2}=\frac{72}5.$$

$$5\times\frac{72}5=72.$$

Solutions · 1