Question

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

Answer 1

Correct Answer: 3

The slopes of \(PQ\) and \(QR\) satisfy:

\( m_{PQ} \cdot m_{QR} = -1. \)

From the given coordinates:

\( m_{PQ} = \frac{10-2}{9+3} = \frac{8}{12} = \frac{2}{3}, \quad m_{QR} = \frac{10-4}{9-\alpha}. \)

Using \( m_{PQ} \cdot m_{QR} = -1 \):

\( \frac{2}{3} \cdot \frac{6}{9-\alpha} = -1 \implies \alpha = 13. \)

Next, calculate the equation of \(QS\):

\( y - 10 = -\frac{4}{7}(x - 9) \implies 4x + 7y = 106 \quad (1). \)

Similarly, the equation of \(RS\) is:

\( y - 4 = -8(x - 13) \implies 8x + y = 108 \quad (2). \)

Solve equations (1) and (2):

\( x = \frac{25}{2}, \quad y = 8. \)

Substituting into \(2x - ky = 1\):

\( 2 \cdot \frac{25}{2} - 8k = 1 \implies 25 - 8k = 1 \implies k = 3. \)

Answer 2

The correct answer is 3.

$m_{PQ}\cdot m_{QR}=-1$
$\Rightarrow \frac{10-2}{9+3}\times \frac{10-4}{9-\alpha }=-1\Rightarrow \alpha =13$
$m_{OP}\cdot m_{QS}=-1\Rightarrow m_{QS}=-\frac{4}{7}$ Equation of QS
$y-10=-\frac{4}{7}(x-9)$
$\Rightarrow 4x+7y=106\dots .(1)$
$m_{OR}\cdot m_{RS}=-1\Rightarrow m_{RS}=-8$
Equation of RS
$y-4=-8(x-13)$
$\Rightarrow 8x+y=\mathrm{108...}$(2)
Solving eq. (1) & (2)
$x_1=\frac{25}{2}{y}_1=8$
$S\left(x_1,y_1\right)\text{lies on}2x-ky=1$
$25-8k=1$
$\Rightarrow 8k=24$
$\Rightarrow k=3$