JEE Mathematics 2023 Question with Solution

Given: Points $P(-3,2), Q(9,10)$ and $R(\alpha,4)$ lie on a circle with $PR$ as diameter. Tangents at $Q$ and $R$ meet at $S$.

Find: The value of $k$ if $S$ lies on $2x-ky=1$.

Since $PR$ is a diameter, the angle subtended by it at $Q$ is a right angle. Therefore,

$$m_{PQ} \cdot m_{QR} = -1$$

Now,

$$m_{PQ} = \frac{10-2}{9-(-3)} = \frac{8}{12} = \frac{2}{3}, \qquad m_{QR} = \frac{10-4}{9-\alpha} = \frac{6}{9-\alpha}$$

So,

$$\frac{2}{3} \cdot \frac{6}{9-\alpha} = -1$$ $$\frac{4}{9-\alpha} = -1$$ $$\alpha = 13$$

Thus $R=(13,4)$.

The center of the circle is the midpoint of $P$ and $R$, so

$$O=\left(\frac{-3+13}{2},\frac{2+4}{2}\right)=(5,3)$$

Slope of $OQ$ is

$$m_{OQ}=\frac{10-3}{9-5}=\frac{7}{4}$$

Hence the tangent at $Q$ has slope

$$m_{QS}=-\frac{4}{7}$$

Therefore, equation of $QS$ is

$$y-10=-\frac{4}{7}(x-9)$$ $$4x+7y=106 \qquad (1)$$

Similarly, slope of $OR$ is

$$m_{OR}=\frac{4-3}{13-5}=\frac{1}{8}$$

So the tangent at $R$ has slope

$$m_{RS}=-8$$

Thus equation of $RS$ is

$$y-4=-8(x-13)$$ $$8x+y=108 \qquad (2)$$

Solving $(1)$ and $(2)$,

$$x=\frac{25}{2}, \qquad y=8$$

So $S\left(\frac{25}{2},8\right)$.

Since $S$ lies on $2x-ky=1$,

$$2\left(\frac{25}{2}\right)-k(8)=1$$ $$25-8k=1$$ $$8k=24$$ $$k=3$$

Therefore, the value of $k$ is $3$.

From the figure, the radius to a point of tangency is perpendicular to the tangent.

So,

$$OQ \perp QS \quad \text{and} \quad OR \perp RS$$

Once $\alpha=13$ is obtained from the right angle at $Q$, the midpoint of diameter $PR$ gives the center $O(5,3)$.

Then,

$$m_{OQ}=\frac{10-3}{9-5}=\frac{7}{4}$$

Hence,

$$m_{QS}=-\frac{4}{7}$$

Using point $Q(9,10)$,

$$y-10=-\frac{4}{7}(x-9)$$

which gives

$$4x+7y=106$$

Also,

$$m_{OR}=\frac{4-3}{13-5}=\frac{1}{8}$$

Hence,

$$m_{RS}=-8$$

Using point $R(13,4)$,

$$y-4=-8(x-13)$$

which gives

$$8x+y=108$$

Solving these two tangent equations gives

$$S\left(\frac{25}{2},8\right)$$

Substitute in $2x-ky=1$:

$$25-8k=1$$

So,

$$k=3$$

Therefore, the correct answer is $3$.