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

Let $S ≡ \left({x}_{1}, {y}_{1}\right)$

Given:

$P ≡ \left(- 3, 2\right)Q ≡ \left(9, 10\right)R ≡ \left(\alpha, 4\right)$

Since, $PQ ⊥ RQ$, so

${m}_{PQ} \times {m}_{QR} = - 1$

$⇒ \left(\dfrac{10 - 2}{9 + 3}\right) \times \left(\dfrac{10 - 4}{9 - \alpha}\right) = - 1$

$⇒ \left(\dfrac{1}{9 - \alpha}\right) = - \dfrac{1}{4}$

$⇒ \alpha = 13$

So, $R ≡ \left(13, 4\right)$

So, centre of circle is $O ≡ \left(\dfrac{13 - 3}{2}, \dfrac{4 + 2}{2}\right) ≡ \left(5, 3\right)$

Also,

${m}_{OQ} \cdot {m}_{QS} = - 1$

$⇒ \left(\dfrac{3 - 10}{5 - 9}\right){m}_{QS} = - 1$

$⇒ {m}_{QS} = - \dfrac{4}{7}$

Equation of $QS$ is

$y - 10 = - \dfrac{4}{7}\left(x - 9\right)$

$⇒ 4x + 7y = 106 \ldots \left(1\right)$

Also,

${m}_{OR} \cdot {m}_{RS} = - 1$

$⇒ \left(\dfrac{4 - 3}{13 - 5}\right){m}_{RS} = - 1$

$⇒ {m}_{RS} = - 8$

Equation of $RS$

$y - 4 = - 8\left(x - 13\right)$

$⇒ 8x + y = 108 \ldots \left(2\right)$

Solving equations \left(1\right) \text{ and } \left(2\right)$, we get</p><p>${x}_{1} = \dfrac{25}{2}; {y}_{1} = 8$</p><p>Since,$S\left({x}_{1}, {y}_{1}\right)$lies on$2x - ky = 1$</p><p>$25 - 8k = 1$</p><p>$⇒ 8k = 24$</p><p>$⇒ k = 3$