Question

The area of the triangle formed by the positive x-axis and the normal and the tangent to the circle $x^2+y^2=4$ at $(1,\sqrt{3})$ is

• A. $2\sqrt{3}$
• B. $3\sqrt{2}$
• C. $\sqrt{6}$
• D. None of these

Answer 1

The correct option is A $2\sqrt{3}$

Let A (1, $\sqrt{3}$) and the triangle formed be OAB where O is origin.
B is the point of intersection of the tangent with the X axis.
The tangent of the circle $x^2+y^2=a^2$ at $(a\cos \theta ,a\sin \theta )$ is $x\cos \theta +y\sin \theta =a.$
So the tangent at A $(1,\sqrt{3})$ is $\sqrt{3}y+x=4$ which intersects $x$ axis at $B(4,0).$
The area af the triangle OAB is $2\sqrt{3}$ sq.units.