If the tangents at the points P and Q on the circle $\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{x}+\mathrm{y}=5$ meet at the point $\mathrm{R}\left(\frac{9}{4}, 2\right)$, then the area of the triangle PQR is

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Circle

If the tangents at the points P and Q on the circle

x^{2} + y^{2} - 2 x + y = 5

meet at the point

R (\frac{9}{4}, 2)

, then the area of the triangle PQR is

Answer: (D)

The circle

x^{2} + y^{2} - 2 x + y - 5 = 0

has centre

C (1, - \frac{1}{2})

and radius

R = 1 + \frac{1}{4} + 5 = \frac{5}{2}

. The square of the length of the tangent from

R (\frac{9}{4}, 2)

t^{2} = (\frac{9}{4})^{2} + 2^{2} - 2 \cdot \frac{9}{4} + 2 - 5 = \frac{25}{16}

, so

P R = QR = t = \frac{5}{4}

. The distance

C R

satisfies

C R^{2} = R^{2} + t^{2} = \frac{25}{4} + \frac{25}{16} = \frac{125}{16}

. If

M

is the midpoint of the chord of contact

P Q

, then

P M = \frac{R t}{C R}

and the distance of

R

from

P Q

R M = \frac{t ^{2}}{C R}

. Hence the area of

△ P QR

\frac{1}{2} P Q \cdot R M = \frac{R t ^{3}}{R ^{2} + t ^{2}} = \frac{\frac{5}{2} \cdot \frac{125}{64}}{\frac{125}{16}} = \frac{5}{8}

. The area of triangle

P QR

\frac{5}{8}

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Single correct option