JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Circle

Let C be the centre of the circle

x^{2} + y^{2} - x + 2 y = \frac{11}{4}

and P be a point on the circle. A line passes through the point C , makes an angle of

\frac{π}{4}

with the line CP and intersects the circle at the points Q and R . Then the area of the triangle PQR (in unit

^{2}

) is :

Answer: (B)

Solution

The circle

x^{2} + y^{2} - x + 2 y = \frac{11}{4}

can be rewritten as

(x - \frac{1}{2})^{2} + (y + 1)^{2} = 4

, with center

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

and radius

r = 2

. The line QR passes through C and makes an angle

\frac{π}{4}

with CP. Since C is the center,

C P = C Q = C R = 2

. The angle

∠ P C Q = ∠ P C R = \frac{π}{4}

, so

∠ QC R = \frac{π}{2}

. In triangle PQR, using the formula for the area with two sides and included angle: Area

_{△ P QR} = \frac{1}{2} \cdot P Q \cdot P R \cdot sin (∠ QP R)

. By geometry,

P Q = P R = 22

(using the law of cosines in isosceles triangles PCQ and PCR). The angle

∠ QP R = \frac{π}{4}

. Area

= \frac{1}{2} \cdot 22 \cdot 22 \cdot sin (\frac{π}{4}) = \frac{1}{2} \cdot 8 \cdot \frac{1}{2} = 22

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Single correct option