Question

As shown in the figure, circle \(C_1\) with center \(O_1\) and radius \(r_1\) touches the square \(VWXY\) at points \(P\) and \(Q\) while circle \(C_2\) with center \(O_2\) and radius \(r_2\) touches the square \(VWXY\) at points \(R\) and \(S\). The two circles touch each other at \(T\).

Given \(r_1 = 1\) cm and \(\overline{VY} = \overline{VW} = 4\) cm, \(r_2 =\) ________ cm.

• A. \(4 - 3\sqrt{2}\)
• B. \(1 + 2\sqrt{2}\)
• C. \(7 - 4\sqrt{2}\)
• D. \(5 + 3\sqrt{2}\)

Hint:

Place the square on coordinates so \(O_1\) sits at distance \(r_1\) from two adjacent sides and \(O_2\) sits at distance \(r_2\) from the other two adjacent sides.
Use the fact that the circles touch, so the distance between \(O_1\) and \(O_2\) equals \(r_1+r_2\), and solve the resulting equation for \(r_2\).

Answer 1

The Correct Option is C

Step 1: Use the diagonal of the square instead of separate coordinates.
Both centers $O_1$ and $O_2$ lie on the same diagonal of the square, the one running from corner $W$ to corner $Y$, because each center is equally far from its two tangent sides, which places it exactly on the line that bisects that corner's right angle. So the whole problem can be solved by measuring distances along this one diagonal.

Step 2: Find the length of the diagonal.
The square has side 4 cm, so its diagonal $WY$ has length:
\[ WY = 4\sqrt{2} \]

Step 3: Find how far each center sits from its nearest corner along the diagonal.
Center $O_1$ sits in the corner at $W$, at perpendicular distance $r_1 = 1$ from each of the two sides meeting at $W$. Since a corner is a right angle, the straight-line distance from $W$ to $O_1$ along the diagonal is $r_1\sqrt{2} = \sqrt{2}$. By the same reasoning, the distance from $Y$ to $O_2$ along the diagonal is $r_2\sqrt{2}$.

Step 4: Write the whole diagonal as three pieces.
Moving from $W$ to $Y$ along the diagonal, the length splits into the distance from $W$ to $O_1$, plus the distance from $O_1$ to $O_2$, which is $r_1+r_2$ since the circles touch, plus the distance from $O_2$ to $Y$:
\[ WY = \sqrt{2}\,r_1 + (r_1+r_2) + \sqrt{2}\,r_2 \]
Substitute $WY = 4\sqrt{2}$ and $r_1 = 1$:
\[ 4\sqrt{2} = \sqrt{2} + (1+r_2) + \sqrt{2}\,r_2 \]

Step 5: Solve for $r_2$.
\[ 4\sqrt{2} - \sqrt{2} - 1 = r_2 + \sqrt{2}\,r_2 \]
\[ 3\sqrt{2} - 1 = r_2(1+\sqrt{2}) \]
\[ r_2 = \frac{3\sqrt{2}-1}{1+\sqrt{2}} \times \frac{\sqrt{2}-1}{\sqrt{2}-1} = \frac{6-3\sqrt{2}-\sqrt{2}+1}{1} = 7-4\sqrt{2} \]

Final Answer:
Measuring along the square's diagonal instead of using two separate coordinates gives the same value, $r_2 = 7-4\sqrt{2}$ cm, confirming option (C). \[ \boxed{r_2 = 7-4\sqrt{2}\text{ cm}} \]