Question

$AD$ is a diameter of a circle. Two more circles pass through $A$ and intersect $AD$ in $B$ and $C$ respectively, such that $AB$ and $AC$ are diameters of these circle and $AD>AC>AB$. If the circumference of the middle circle is average of the circumference of the other two, Then given $AB=4$ units and $CD=2$ units. Then the area, in sq. units of the largest circle is

• A. $128\pi$
• B. $64\pi$
• C. $48\pi$
• D. $16\pi$

Answer 1

The correct option is D $16\pi$

Given- AD is the diameter of a circle with $C_3$ as circumference. Two other circles of circumferences $C_1$ & $C_2$ pass through $A$ such that that they intersect $AD$ at $B$ & $C$ respectively. $AD>AC>AB.AB=4units,CD=2units.C_2$ is the average of $C_1\&C_3$.

To find out- $ar.C_3=?$

Solution- Let $BC=x.\therefore AD=(4+x+2)units=(6+x)units.So{C}_3=\pi (6+x)units$........(i).

Similarly, $AC=(4+x)unitsSo{C}_2=\pi (4+x)units........\left(ii\right).$ Also $AB=4units,$

$\therefore C_1=4\pi units$.........(iii)

Now, by the given condition, $C_2=\frac{C_1+C_3}{2}⟹\pi (4+x)=\frac{4\pi +\pi (6+x)}{2}$ (from i, ii \& iii)

$⟹x=2units$. $\therefore AD=(6+2)units=8units.i.erad.C_3=(8÷2)units=4units.$

So $ar.C_3=\pi 4^2$ sq.units$=16\pi$ sq.units

Ans- Option D.