Let C be the centre of the circle x2+y2-x+2y=114 and P be a point on the circle. A line passes through the point C, makes an angle of π4with the line CP and intersects the circle at the points Q and R. Then the area of the triangle PQR (in unit2)is :

x2+y2-x+2y=114 x-122+y+12=22 Or ∆ PQR PR=QRsinπ8⇒PR=4 sinπ8 PQ=QR cos 2212 ⇒PQ=4 cos π8 As ∆ PQR=12PR×PQ =124 sinπ84cosπ8 =4 sinn π4=42=22

Let C be the centre of the circle x2+y2-x+2y=114 and P be a point on the circle. A line passes through the point C, makes an angle of π4with the line CP and intersects the circle at the points Q and R. Then the area of the triangle PQR (in unit2)is :

x2+y2-x+2y=114 x-122+y+12=22 Or ∆ PQR PR=QRsinπ8⇒PR=4 sinπ8 PQ=QR cos 2212 ⇒PQ=4 cos π8 As ∆ PQR=12PR×PQ =124 sinπ84cosπ8 =4 sinn π4=42=22

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

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 unit²)
is :

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$8 \cos (\frac{π}{8})$

$8 \sin (\frac{π}{8})$

$2 \sqrt{2}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$x^{2} + y^{2} - x + 2 y = \frac{11}{4} {(x - \frac{1}{2})}^{2} + {(y + 1)}^{2} = {(2)}^{2} O r ∆ P Q R P R = Q R \sin \frac{π}{8}$

$\Rightarrow P R = 4 \sin \frac{π}{8} P Q = Q R \cos 22 \frac{1}{2} \Rightarrow P Q = 4 \cos \frac{π}{8} A s ∆ P Q R = \frac{1}{2} P R \times P Q = \frac{1}{2} (4^{} \sin \frac{π}{8}) (4 \cos \frac{π}{8}) = 4 \sin n \frac{π}{4} = \frac{4}{\sqrt{2}} = 2 \sqrt{2}$