Let the tangents at two points [A] and [B] on the circle [x^2+y^2-4 x+3=0] meet at origin …

Question

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

Accepted Answer

${x^2} + {y^2} - 4x + 3 = 0$

$$ \Rightarrow {(x - 2)^2} + {y^2} = 1$$

$$AO = \sqrt {{{(OC)}^2} - {{(AC)}^2}} $$

$$ = \sqrt {4 - 1}  = \sqrt 3 $$

$$\sin \theta  = {1 \over 2} \Rightarrow \theta  = {\pi  \over 6}$$

Also, $AO = BO$

Area of $$\Delta OAB = {1 \over 2}\,.\,OA\,.\,OB\sin 60^\circ $$

$$ = {1 \over 2} \times \sqrt 3 \,.\,\sqrt 3 \,.\,{{\sqrt 3 } \over 2} = {{3\sqrt 3 } \over 4}$$

Solutions · 1