Let the tangents at two points A and B on the circle x^2+y^2-4x+3=0 meet at origin O(0,0).

Question

Let the tangents at two points $A$ and $B$ on the circle $x^2+y^2-4x+3=0$ meet at origin $O(0,0)$. Then the area of the triangle $OAB$ is:

Accepted Answer

Circle: $(x-2)^2+y^2=1$, centre $C=(2,0)$, $r=1$. $OC=2$.

$$OA=\sqrt{OC^2-r^2}=\sqrt{4-1}=\sqrt3,\quad \sin\theta=\frac r{OC}=\frac12\Rightarrow\theta=\frac\pi6,$$
so $\angle AOB=\frac\pi3$. Since $OA=OB=\sqrt3$:
$$\text{Area}=\frac12\cdot\sqrt3\cdot\sqrt3\cdot\sin60^\circ=\frac32\cdot\frac{\sqrt3}2=\frac{3\sqrt3}4.$$

Solutions · 1