JEE Main 2019 — Circle Question with Solution

Equation of tangent to the circle x² + y² = 4 at point ( $$\sqrt{3}$$, 1) is

$$\sqrt{3}$$x + y = 4

Slope of this tangent (m) = - $$\sqrt{3}$$

$$\therefore$$ Slope of the normal at point ( $$\sqrt{3}$$, 1) is = $$\frac{1}{\sqrt{3}}$$

$$\therefore$$ Equation of normal at point ( $$\sqrt{3}$$, 1),

y - 1 = $$\frac{1}{\sqrt{3}}\left(x-\sqrt{3}\right)$$

$$\Rightarrow$$ x - $$\sqrt{3}$$y = 0

So normal passes through the center of the axis.

Tangent cut the x-axis at point A when y = 0,

So the x coordinate of the point A is

$$\sqrt{3}$$x + 0 = 4

$$\Rightarrow$$ x = $$\frac{4}{\sqrt{3}}$$

$$\therefore$$ Coordinate of A = $$\left(\frac{4}{\sqrt{3}},0\right)$$

Area of triangle OAB,

$$\Delta$$ = {1 \over 2}\left| {\matrix{ 0 & 0 & 1 \cr {{4 \over {\sqrt 3 }}} & 0 & 1 \cr {\sqrt 3 } & 1 & 1 \cr } } \right|

= $$\frac{1}{2}\left(\frac{4}{\sqrt{3}}\right)$$

= $$\frac{2}{\sqrt{3}}$$