The tangent and the normal lines at the point | vedclass

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(D) Let the circle be $x^2 + y^2 = 4$,with center $O(0, 0)$ and radius $r = 2$. The point $P$ is $(\sqrt{3}, 1)$.The normal at $P$ is the line passing

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let the circle be $x^2 + y^2 = 4$,with center $O(0, 0)$ and radius $r = 2$. The point $P$ is $(\sqrt{3}, 1)$.The normal at $P$ is the line passing through $O(0, 0)$ and $P(\sqrt{3}, 1)$. Its equation is $y = \frac{1}{\sqrt{3}}x$,or $x - \sqrt{3}y = 0$.The tangent at $P(\sqrt{3}, 1)$ to the circle $x^2 + y^2 = 4$ is given by $xx_1 + yy_1 = r^2$,which is $\sqrt{3}x + y = 4$.The tangent intersects the $x$-axis at point $Q$. Setting $y = 0$ in the tangent equation,we get $\sqrt{3}x = 4$,so $x = \frac{4}{\sqrt{3}}$. Thus,$Q = (\frac{4}{\sqrt{3}}, 0)$.The triangle is formed by the origin $O(0, 0)$,the point $P(\sqrt{3}, 1)$,and the point $Q(\frac{4}{\sqrt{3}}, 0)$.The area of triangle $OPQ$ is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$.Base $OQ = \frac{4}{\sqrt{3}}$ and height (the $y$-coordinate of $P$) $= 1$.Area $= \frac{1}{2} 	imes \frac{4}{\sqrt{3}} 	imes 1 = \frac{2}{\sqrt{3}}$ square units.

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

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