Question

If two circles are such that the centre of one lies on the circumference of the other, then the ratio of the common chord of the two circles to the radius of any one of the circles is

• A. $2:1$
• B. $\sqrt{3}:1$
• C. $\sqrt{5}:1$
• D. $4:1$

Answer 1

The correct option is B $\sqrt{3}:1$

$C_1{C}_2=r_1=r_2\therefore C_1\&C_{2}areidenticalcircles.In△A{C}_1{C}_2\&△C_1{C}_{2}B:\Rightarrow C_1{C}_2=C_1{C}_2=r_1=r_2\left(common\right)A{C}_1=C_{1}BA{C}_2=C_{2}B\therefore △A{C}_1{C}_2\cong △C_1{C}_{2}BbyS.S.S.AlsoA{C}_1=A{C}_2=C_1{C}_2=r_1=r_{2}and{C}_1{C}_2=C_{1}B=C_{2}B=r_1=r_{2}andAO=OB=d\left(let\right)\therefore △A{C}_1{C}_2\&△C_1{C}_{2}Bareequilateraltriangles.\therefore \angle B{C}_1{C}_2=\angle A{C}_1{C}_2={60}^{0}In△A{C}_{1}O\&△O{C}_{1}B\Rightarrow △O{C}_1=O{C}_1\left(common\right)\angle O{C}_{1}A=\angle O{C}_{1}BbyS.A.S.\Rightarrow \angle AO{C}_1=\angle C_{1}OB=x\left(let\right).Sumofanglesonastraightline={180}^0\Rightarrow 2x={180}^0\Rightarrow x={90}^{0}In△AO{C}_1\&△AO{C}_2\Rightarrow \angle AO{C}_1=\angle AO{C}_2={90}^0\Rightarrow A{C}_1=A{C}_2=r_1=r_{2}AO=AO\left(common\right)Hence,△AO{C}_1\cong △AO{C}_{2}byR.H.S.Hence,C_{1}O=C_{2}O=\frac{r_1}{2}=\frac{r_2}{2}.In△A{C}_{1}O:A{C}_1^2=A{O}^2+C_1{O}^2\Rightarrow r_1^2=d^2+\frac{r_1^2}{4}\Rightarrow \frac{{3r}^2}{4}=d^2\Rightarrow (\frac{d}{r}{)}^2=\frac{3}{4}\Rightarrow \frac{d}{r}=\frac{\sqrt{3}}{2}.\therefore \frac{d}{r}=\sqrt{3}.andhence\frac{AB}{radius}=\sqrt{3}:1$