Question

Let $S_1$ and $S_2$ be two circles touching externally and having radius as $2$ and $3$ respectively. $S_1$ and $S_2$ touch a variable circle $S_3$ internally at $A$ and $B$ respectively. If the tangents to $S_3$ at $A$ and $B$ meet at $T$ and $TA=4\text{units}$, then which of the following is/are correct?
(Here, $C_i$ represents the centre of circle $S_i$)

• A. The radius of circle $S_3$ is $8$ units.
• B. The area of circle circumscribing $△TAB$ is $20\pi$ sq. units.
• C. $C_3{C}_1+C_3{C}_2=5$
• D. $C_3{C}_1–C_3{C}_2=1$

Answer 1

Answer: (A), (B), (D)

The correct options are
A The radius of circle $S_3$ is $8$ units.
B The area of circle circumscribing $△TAB$ is $20\pi$ sq. units.
D $C_3{C}_1–C_3{C}_2=1$

$AT$ is a radical axis of $S_1$ and $S_3$
$BT$ is a radical axis of $S_2$ and $S_3$
$\therefore T$ is radical centre.
So, common tangent of $S_1$ and $S_2$ should pass through $T$.

$TA=TB=TD=4$
In $△C_{1}AT$
$\tan \frac{{\theta }_1}{2}=\frac{2}{4}\Rightarrow \tan \frac{{\theta }_1}{2}=\frac{1}{2}$
In $△C_{2}BT$
$\tan \frac{{\theta }_2}{2}=\frac{3}{4}$

In $△C_{3}AT$
$\tan \frac{{\theta }_1+{\theta }_2}{2}=\frac{r_3}{4}\Rightarrow \frac{r_3}{4}=\frac{\frac{1}{2}+\frac{3}{4}}{1-\frac{1}{2}\cdot \frac{3}{4}}=2$
$\therefore r_3=8$ units

Circumference of $△TAB$ passes through $C_3$ and having $T{C}_3$ as diameter.

$\cos \left(\frac{{\theta }_1+{\theta }_2}{2}\right)=\frac{4}{C_{3}T}\Rightarrow \frac{1}{\sqrt{5}}=\frac{4}{C_{3}T}\Rightarrow C_{3}T=4\sqrt{5}$
Area $=\pi {\left(\frac{T{C}_3}{2}\right)}^2=\pi \times (2\sqrt{5}{)}^2$
$\therefore$ Area $=20\pi$ sq. units

$C_3{C}_1=r_3-r_1{C}_3{C}_2=r_3-r_2\therefore C_3{C}_1-C_3{C}_2=r_2-r_1=1$