Problem :

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

My approach :

$x^2+y^2 = 4 $

$2x +2y \frac{dy}{dx}=0$

$\Rightarrow \frac{dy}{dx}=\frac{-x}{y} $

at point $(1,\sqrt{3})$ slope of tangent

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

So slope of normal = $\sqrt{3}$

Now how to find the area of the triangle $\Delta$ Please guide will be of great help thanks.

## Best Answer

The slope of the radius to $\;(1,\sqrt3)\;$ is $\;\sqrt3\;$, which is exactly the same as the slope of the normal (and observe it passes through the origin = the circle's center), so the slope of the tangent is $\;-\frac1{\sqrt3}\;$, and thus the tangent line is:

$$y-\sqrt3=-\frac1{\sqrt3}(x-1)\;,\;\;\text{which intersects the hotizonal axis at}\;\;(4,0)$$

and thus the triangle's vertices are

$$(0,0)\,,\;\;(1,\sqrt3)\,,\;\;(4,0)\implies S_\Delta=2\sqrt3$$

No need of differential calculus.