Question

The area of the triangle formed by joining the origin to the points of intersection of the line $x\sqrt{5}+2y=3\sqrt{5}$ and

circle $x^2+y^2=10$ is

• A. 3
• B. 4
• C. 5
• D. 6

Answer 1

The correct option is C

5

Length of perpendicular from origin to the line x\sqrt{5} + 2y = 3\sqrt{5} is

OL = $\frac{3\sqrt{5}}{(\sqrt{5}{)}^2+2^2}=\frac{3\sqrt{5}}{\sqrt{9}}=\sqrt{5}$

Radius of the given circle = $\sqrt{10}$ = OQ = OP

$PQ=2QL=2\sqrt{O{Q}^2-O{L}^2}=2\sqrt{10-5}=2\sqrt{5}$

Thus area of \triangle OPQ = $\frac{1}{2}\times PQ\times OL=\frac{1}{2}\times 2\sqrt{5}\times \sqrt{5}=5$.