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

If the circle $x^2 + y^2 - 6x - 8y + (25 - a^2) = 0$ touches the $x$-axis,then $a$ equals

The equation of the circle with $(1,1)$ as centre and which cuts a chord of length $4 \sqrt{2}$ units on the line $x+y+1=0$ is

If $h, k, p, q \neq 0$ and the circles $x^2+y^2+2hx+2ky=0$ and $x^2+y^2+2px+2qy=0$ touch each other at the origin,then $hq-pk-\frac{hq}{pk}$ is equal to

Let $x-y=0$ and $x+y=1$ be two perpendicular diameters of a circle of radius $R$. The circle will pass through the origin if $R$ is equal to

If $P(2,3)$ and $Q(-1,2)$ are conjugate points with respect to the circle $x^2+y^2+2gx+3y-2=0$,then the radius of the circle is

Vedclass Test Series

Exam Paper Generator

Online Exam Module