A pair of tangents are drawn to a unit circle | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the circle be $x^2 + y^2 = 1$,so the radius $r = 1$. Let the tangents intersect at point $A$. The angle between the tangents is $60^o$,so the

Vedclass · Accepted Answer

$\sqrt{3} - \frac{\pi}{3}$

Vedclass · Answer

(B) Let the circle be $x^2 + y^2 = 1$,so the radius $r = 1$. Let the tangents intersect at point $A$. The angle between the tangents is $60^o$,so the angle between each tangent and the line joining the origin to $A$ is $30^o$. Let $O$ be the origin and $P$ be the point of tangency. In $	riangle OPA$,$\angle OPA = 90^o$ and $\angle OAP = 30^o$. Thus,$PA = r \cot(30^o) = 1 \cdot \sqrt{3} = \sqrt{3}$. The area of the quadrilateral formed by the origin,the two points of tangency,and $A$ is $2 	imes (	ext{Area of } 	riangle OPA) = 2 	imes (\frac{1}{2} \cdot PA \cdot r) = 2 	imes (\frac{1}{2} \cdot \sqrt{3} \cdot 1) = \sqrt{3}$. The angle subtended by the two points of tangency at the center is $180^o - 60^o = 120^o = \frac{2\pi}{3}$ radians. The area of the circular sector is $\frac{1}{2} r^2 	heta = \frac{1}{2} \cdot 1^2 \cdot \frac{2\pi}{3} = \frac{\pi}{3}$. The area enclosed by the tangents and the arc is the area of the quadrilateral minus the area of the sector: $\sqrt{3} - \frac{\pi}{3}$.

$A$ pair of tangents are drawn to a unit circle with centre at the origin and these tangents intersect at a point $A$ enclosing an angle of $60^o$. The area enclosed by these tangents and the arc of the circle is

Similar Questions

$A$ rhombus is inscribed in the region common to the two circles $x^2 + y^2 - 4x - 12 = 0$ and $x^2 + y^2 + 4x - 12 = 0$,with two of its vertices on the line joining the centres of the circles. The area of the rhombus is:

The values of the constant term in the equation of a circle passing through $(1, 2)$ and $(3, 4)$ and touching the line $3x + y - 3 = 0$ are

If the common tangent to the parabolas $y^{2}=4x$ and $x^{2}=4y$ also touches the circle $x^{2}+y^{2}=c^{2}$,then $c$ is equal to

Let $y=mx+c, m>0$ be the focal chord of $y^{2}=-64x$,which is tangent to $(x+10)^{2}+y^{2}=4$. Then,the value of $4\sqrt{2}(m+c)$ is equal to $.....$

$A$ circle $S \equiv x^2+y^2+2gx+2fy+6=0$ cuts another circle $x^2+y^2-6x-6y-6=0$ orthogonally. If the angle between the circles $S=0$ and $x^2+y^2+6x+6y+2=0$ is $60^{\circ}$,then the radius of the circle $S=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module