Let S_1, S_2, and S_3 be three circles of uni | vedclass

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

Question

(B) Let the centers of the three circles form an equilateral triangle of side length $2$. The distance from the center of a circle to the side of the

Vedclass · Accepted Answer

$2(1+\frac{1}{\sqrt{3}})$

Vedclass · Answer

(B) Let the centers of the three circles form an equilateral triangle of side length $2$. The distance from the center of a circle to the side of the large triangle $ABC$ is $1$. The angle of the triangle $ABC$ is $60^{\circ}$ because the circles are arranged symmetrically.From the geometry,the base $BC$ consists of the distance between the centers of the two bottom circles $(2)$ plus the horizontal projections of the distances from the centers to the vertices $B$ and $C$. Since the radius is $1$ and the angle is $60^{\circ}$,the distance from the projection of the center to the vertex is $1 \cdot \cot(30^{\circ}) = \sqrt{3}$.Thus,$BC = \sqrt{3} + 2 + \sqrt{3} = 2(1+\sqrt{3})$.Since $ABC$ is an equilateral triangle with side $a = 2(1+\sqrt{3})$,the circumradius $R$ is given by $R = \frac{a}{2 \sin(60^{\circ})} = \frac{2(1+\sqrt{3})}{2 \cdot \frac{\sqrt{3}}{2}} = \frac{2(1+\sqrt{3})}{\sqrt{3}} = 2(\frac{1}{\sqrt{3}} + 1)$.

Let $S_1, S_2,$ and $S_3$ be three circles of unit radius which touch each other externally. The common tangents to each pair of circles are drawn and extended so that they intersect and form a triangle $ABC$ with circumradius $R$. Then $R$ is equal to

Similar Questions

The equation of a circle touching the curve $|x - 1| + |y - 4| = 6$ is

$A$ square is inscribed in the circle $x^2+y^2-2x+4y-93=0$ with its sides parallel to the coordinate axes. Which of the following can be one of the vertices of the square?

If the circle $x^2+y^2+2gx+2fy+c=0$ $(c>0)$ touches both the coordinate axes and lies in the third quadrant,then the length of the chord intercepted by the circle on the line $x+y+\sqrt{c}=0$ is

From a point $P$ on the line $4x - 3y = 6$,two tangents are drawn to the circle $x^2 + y^2 - 6x - 4y + 4 = 0$. If the angle between these tangents is $\tan^{-1}\left(\frac{24}{7}\right)$,then $P$ can be:

For the two circles $x^2+y^2=16$ and $x^2+y^2-2y=0$, there is/are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module