Points P(-3,2), Q(9,10), and R(,4) lie on a circle C with PR as its diameter.

Question

Points $P(-3,2)$, $Q(9,10)$, and $R(\alpha,4)$ lie on a circle $C$ with $PR$ as its diameter. The tangents to $C$ at $Q$ and $R$ intersect at $S$. If $S$ lies on the line $2x-ky=1$, then $k$ is equal to \_\_\_\_\_\_\_.

Answer: 3

Accepted Answer

Since $PR$ is a diameter, $\angle PQR=90^\circ$, so $PQ\perp QR$:
$$m_{PQ}=\frac{10-2}{9+3}=\frac23,\qquad m_{PQ}\cdot m_{QR}=-1\Rightarrow \frac23\cdot\frac{6}{9-\alpha}=-1\Rightarrow \alpha=13.$$

So $R=(13,4)$; centre $O=\left(\dfrac{-3+13}{2},\dfrac{2+4}{2}\right)=(5,3)$.

Tangent at $Q$: $m_{OQ}=\dfrac{10-3}{9-5}=\dfrac74$, so tangent slope $=-\dfrac47$: $4x+7y=106$.
Tangent at $R$: $m_{OR}=\dfrac{4-3}{13-5}=\dfrac18$, so tangent slope $=-8$: $8x+y=108$.

Solving: $x=\dfrac{25}{2},\ y=8$, so $S=\left(\dfrac{25}{2},8\right)$. On $2x-ky=1$: $25-8k=1\Rightarrow k=3.$

Solutions · 1