Question

If the tangent and the normal at the point \((\sqrt{3}, 1)\) to the circle \(x^2 + y^2 = 4\), and the X-axis form a triangle, then the area (in sq.units) of this triangle is

• A. \(\frac{1}{\sqrt{2}}\)
• B. \(\frac{2}{\sqrt{3}}\)
• C. \(\frac{4}{\sqrt{3}}\)
• D. \(\frac{1}{3}\) ip Note: The direct calculation gives area \(= \dfrac{2}{\sqrt{3}}\), which does not match any provided option.

Hint:

For circles centred at the origin, the normal at any point always passes through the origin. That makes such triangle-area questions much faster.

Answer 1

The Correct Option is B

Concept:
For the circle \[ x^2+y^2=4, \] the tangent at \((x_1,y_1)\) is: \[ xx_1+yy_1=4 \] The normal at a point on a circle passes through the centre. ip

Step 1: Write the tangent equation.
At the point \[ (\sqrt{3},1), \] the tangent is: \[ \sqrt{3}x + y = 4 \] Its x-intercept is obtained by putting \(y=0\): \[ \sqrt{3}x=4 \Rightarrow x=\frac{4}{\sqrt{3}} \] So tangent meets x-axis at: \[ \left(\frac{4}{\sqrt{3}},0\right) \] ip

Step 2: Write the normal equation.
Since the normal passes through the origin and \((\sqrt{3},1)\), its slope is: \[ \frac{1}{\sqrt{3}} \] So the normal is: \[ y=\frac{x}{\sqrt{3}} \] This meets x-axis at: \[ (0,0) \] ip

Step 3: Find the area of the triangle formed.
The triangle is formed by:
• the origin \((0,0)\),
• the tangent x-intercept \(\left(\frac{4}{\sqrt{3}},0\right)\),
• the point \((\sqrt{3},1)\) e} Base on x-axis: \[ \frac{4}{\sqrt{3}} \] Height from \((\sqrt{3},1)\) to x-axis: \[ 1 \] So area: \[ \frac12 \cdot \frac{4}{\sqrt{3}} \cdot 1 = \frac{2}{\sqrt{3}} \] ip Thus, the mathematical area is:
\[ \boxed{\frac{2}{\sqrt{3}}} \]