The common chord of two intersecting circles C1 & C2 can be seen

Question

The common chord of two intersecting circles C1 and C2 can be seen from their centres at the angles of \(60^{\circ}\) and \(120^{\circ}\) respectively. Find the ratio of area of the smaller circle to the area of the larger one.

A. \(1:2\)
B. \(1:3\)
C. \(1:4\)
D. \(2:3\)
E. \(1:\sqrt{3}\)

A.  1:2 
B.  1:3 
C.  1:4 
D.  2:3 
E.  1:\sqrt{3}

chetan2u · Answer

Refer the attached figure.

The common chord is BD, with a length R, which is the radius of bigger circle . Why?
AB=AD=radius, so angle ABD=angle ADB
Angle BAD is given as 60, making other two angles ABD and ADB also 60 each.

Method I: SINE rule ( Not in scope of GMAT)

Take triangle BCD. =>  \frac{Sin(\angle BCD)}{BD}=\frac{Sin(\angle CDB)}{BC}

\frac{Sin( 120)}{R}=\frac{Sin(30)}{r}

\frac{\sqrt{3}}{2R}=\frac{1}{2r}

\frac{r}{R}=\sqrt{3}

Ratio of area will be square of ratio of radius. =>  1^2:(\sqrt{3})^2=1:3

Method II: Pythagoras ( GMAT solution)

Take triangle ABC. 
 \angle CAB = \frac{\angle DAB}{2}=30 
 \angle ACB = \frac{\angle DCB}{2}=60

Thus ABC is a 30-60-90 triangle with sides BC:AB:AC in ratio  1:\sqrt{3}:2 . So r:R= 1:\sqrt{3}

Ratio of area will be square of ratio of radius. =>  1^2:(\sqrt{3})^2=1:3

B

The common chord of two intersecting circles C1 & C2 can be seen

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The common chord of two intersecting circles C1 & C2 can be seen