Question

Let $S_1$ and $S_2$ be two circles drawn inside a unit square ABCD, touching each other externally. Further, the circle $S_1$ touches the sides AD and DC; and the circle $S_2$ touches the sides AB and BC. If the area of $S_2$ is twice the area of $S_1$, then the radius of $S_1$ is

• A. $3\sqrt{2} - 4$
• B. $3 - 2\sqrt{2}$
• C. $2 - \sqrt{2}$
• D. $\frac{3}{\sqrt{2}} - 2$

Hint:

Using the diagonal of the square is a great way to verify: the total diagonal of the unit square is $\sqrt{2}$.
The diagonal is also equal to $r_1\sqrt{2} + r_1 + r_2 + r_2\sqrt{2}$, which directly yields $(r_1+r_2)(1+\sqrt{2}) = \sqrt{2}$.

Answer 1

The Correct Option is A

Step 1: Place the square on axes.
Put the unit square with $D = (0, 0)$, $C = (1, 0)$, $B = (1, 1)$, $A = (0, 1)$. Circle $S_1$ touches the two sides meeting at $D$, and $S_2$ touches the two sides meeting at $B$.

Step 2: Locate the two centres.
Let $S_1$ have radius $r_1$; touching the sides at corner $D$, its centre is at $(r_1, r_1)$. Let $S_2$ have radius $r_2$; touching the sides at corner $B$, its centre is at $(1 - r_2, 1 - r_2)$.

Step 3: Use the area relation.
Area of $S_2$ is twice that of $S_1$, so $r_2^2 = 2 r_1^2$, giving $r_2 = \sqrt{2}\, r_1$.

Step 4: Use the external touching condition.
The distance between the centres equals $r_1 + r_2$. Both centres lie on the diagonal, so the distance is $\sqrt{2}\,\big[(1 - r_2) - r_1\big] = \sqrt{2}\,(1 - r_1 - r_2)$. Setting this equal to $r_1 + r_2$: $\sqrt{2}(1 - r_1 - r_2) = r_1 + r_2$.

Step 5: Substitute $r_2 = \sqrt{2}\,r_1$.
Then $r_1 + r_2 = r_1(1 + \sqrt{2})$. The equation becomes $\sqrt{2} = (r_1 + r_2)(1 + \sqrt{2}) = r_1(1 + \sqrt{2})^2$. Note $(1 + \sqrt{2})^2 = 3 + 2\sqrt{2}$.

Step 6: Solve for $r_1$ and rationalise.
$r_1 = \frac{\sqrt{2}}{3 + 2\sqrt{2}}$. Multiply top and bottom by $(3 - 2\sqrt{2})$: denominator $= 9 - 8 = 1$, numerator $= \sqrt{2}(3 - 2\sqrt{2}) = 3\sqrt{2} - 4$. So $r_1 = 3\sqrt{2} - 4$.
\[ \boxed{3\sqrt{2} - 4} \]