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10-1.Circle and System of Circles
hard
The tangent and the normal lines at the point $(\sqrt 3,1)$ to the circle $x^2 + y^2 = 4$ and the $x -$ axis form a triangle. The area of this triangle (in square units) is
A
$\frac{1}{{\sqrt 3 }}$
B
$\frac{4}{{\sqrt 3 }}$
C
$\frac{1}{3}$
D
$\frac{2}{{\sqrt 3 }}$
(JEE MAIN-2019)
Solution
Slope of $OP = \frac{1}{{\sqrt 3 }}$
Equation of $PQ$ is
$y – 1 = – \sqrt 3 \left( {x – \sqrt 3 } \right)$
$ \Rightarrow y + \sqrt 3 x = 4$
$ \Rightarrow Q\left( {\frac{4}{{\sqrt 3 }},0} \right)$
Area $ = \frac{2}{{\sqrt 3 }}$
Standard 11
Mathematics
