A circle having centre $c_1$, radius $r_1$ = 5 cm is placed against a right angle. Another smaller circle having centre $c_2$, radius $r_2$ is also placed touching the sides of angle and the bigger circle as shown in the figure. Find the radius, $r_2$ in cm, of the smaller circle.
To solve this problem, we will use the concept of tangent circles and geometry related to right-angled triangles.
The larger circle with center \( c_1 \) has a radius \( r_1 = 5 \) cm. The smaller circle with center \( c_2 \) is tangent to both sides of the right angle and the larger circle.
We need to find the radius \( r_2 \) of the smaller circle. Based on the given setup and the figure, the centers of the two circles, \( c_1 \) and \( c_2 \), and the point where the smaller circle touches one of the axes form a right triangle.
Let the side length of the right triangle be \( r_1 + r_2 \) because \( c_1c_2 = r_1 + r_2 \) (as \( c_1 \) to \( c_2 \) is the tangent line from \( c_2 \) to the larger circle).
Then, using the Pythagorean theorem in the right-angle triangle formed, we have:
- The larger circle's center lies \( r_1 \) distance away from both axes, forming the "legs" of the triangle.
- The formula for the distance between the centers (i.e., the hypotenuse) would be \( r_1 + r_2 \).
Thus, applying the Pythagorean theorem:
\(c_1c_2 = \sqrt{(r_1 - r_2)^2 + (r_1 - r_2)^2}\)
Now, solving for \( r_2 \):
\(r_1 + r_2 = \sqrt{2}(r_1 - r_2)\)
Rearranging the terms gives:
\(r_1 + r_2 = \sqrt{2}r_1 - \sqrt{2}r_2\)
Solving for \( r_2 \):
- Collecting like terms: \(r_2 + \sqrt{2}r_2 = \sqrt{2}r_1 - r_1\)
- \(r_2(1 + \sqrt{2}) = r_1(\sqrt{2} - 1)\)
- \(r_2 = \frac{r_1(\sqrt{2} - 1)}{1 + \sqrt{2}}\)
- Substitute \( r_1 = 5 \) cm: \(r_2 = \frac{5(\sqrt{2} - 1)}{1 + \sqrt{2}}\)
- Rationalize the denominator: \(r_2 = 5(\sqrt{2} - 1) \times \frac{1 - \sqrt{2}}{1 - \sqrt{2}} = 5(2 - \sqrt{2} - 1 + 1) = 5(3 - 2\sqrt{2})\)
The radius of the smaller circle is \( r_2 = 5(3 - 2\sqrt{2}) \) cm.
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