If the length $BC=l$, length of arc $AB=l_1$ and length of arc $AC=l_2$, then $l+l_1+l_2=$

Question

Question:

A circle with centre $C_1$ and radius $\frac32$ touches another circle with centre $C_2$ and radius $\frac12$ externally at point $A$. A common tangent touches circle with centre $C_1$ at B and the other circle at C. If the length $BC=l$, length of arc $AB=l_1$ and length of arc $AC=l_2$, then $l+l_1+l_2=$

A) $\frac{5π}6+√3$

B) $\frac{2π}6+√3$

C) $π-√3$

D) none of these

My Attempt:

$l=√{(\frac32+\frac12)^2-(\frac32-\frac12)^2}=√3$

Length of arc $=r\theta$

We have $r$, but not $\theta$.

How to proceed next?

Answer 1

Let $D$ is a point on $C_1B$ such that $CD\parallel C_1C_2$ and $CC_2\perp BC, BC_1\perp BC$ (rad$\perp $tan)

then $BC_1\parallel CC_2$ and $DC_1C_2C$ is a parallelogram

$CD=C_1C_2=r_1+r_2=2$ and $BD=r_1-r_2=1$

Since $C_1B\perp BC$ (rad$\perp $tan)

$BC^2=l^2+BD^2$, $l^2=4-1=3$, $l=\sqrt3$

$\angle BDC=\cos^{-1} (\frac{BD}{CD})=\cos^{-1}(\frac {1}{2})$

then $\angle BDC=60^{\circ}$ so $\angle C_2C_1B=60^{\circ}$ and $l_1=\frac{2\pi r_1}{6}=\frac{2\times \pi \times \frac{3}{2}}{6}= \frac{\pi}{2}$

then $\angle BCD=30^{\circ}$ so $\angle C_1C_2C=120^{\circ}$ and $l_2=\frac{2\pi r_2}{3}=\frac{2\times \pi \times \frac{1}{2}}{3} \frac{\pi}{3}$

$l+l_1+l_2=\sqrt3 + \frac{5\pi}{6}$

Answer 2

Wanted to attempt this one too because I wasn't following all of the steps from the attempted answers. The common tangent between the circles is of length $l$ and forms the base of a rectangle. However, since circle at center $C_1$ has a greater radius, then we can draw a triangle on top of the rectangle.

That rectangle then has the smallest side as $r_1 - r_2 = 1$, and its hypothenuse is $r_1 + r_2 = 2$. The unknown side, which coincides with the base of the rectangle and hence with $l$ is then $\sqrt{2^2 - 1^2} = \sqrt{3}$

As for the arcs, the general formula for the length of the arc is $l_i = r \theta$. So we need to find the angles. It turns out that the triangle that can be drawn on top of the rectanlge, its angle coincide with the angles of $l_1$ and $l_2$. So if you happen to have your triangle with hypothenuse of 2 and sides of 1 and $\sqrt{3}$, then thats the special triangle in which one of the angles is $\pi / 6$ and the other angle is $\pi /3$. If you look for the tangent of the angle $\theta_2$, the angle subtended by the arc on the circle $C_2$ then you get $\tan \frac{1}{\sqrt{3}}$ so that angle is the $\pi/6$ and the other one is the $\pi/3$ one.

However, on the smaller angle, we need to add $\pi/2$ because the angle we got above only covers the angle within the triangle, not the total angle for the arc shown in the picture, so we need to work with $\pi / 6 + \pi /2 = 2\pi/3$

Putting everything together, we can get $$ l + l_1 + l_2 = \sqrt{3} + \frac{3}{2}\frac{\pi}{3} + \frac{1}{2}\frac{2\pi}{3} $$ $$ = \sqrt{3} + \frac{\pi}{2} + \frac{\pi}{3} = \sqrt{3} + \frac{5\pi}{6} $$