Let `ABCD` is a rectangle with `AB=a` & `BC=b` & circle is drawn passing through `A` & `B` and touching side `CD`. Another circle is drawn passing through `B` & `C` and touching side `AD`. Let `r_(1)` & `r_(2)` be the radii of these two circle respectively.
Minimum value of `(r_(1)+r_(2)` equals

Options

A. `5/8 (a-b)`
B. `5/8 (a+b)`
C. `3/8 (a-b)`
D. `3/8 (a+b)`

Explanation

To find the minimum value of \( r_1 + r_2 \) for the given rectangle \( ABCD \) with sides \( AB = a \) and \( BC = b \), we will follow these steps:

### Step 1: Understanding the Geometry
We have a rectangle \( ABCD \). The circle passing through points \( A \) and \( B \) touches side \( CD \), and the circle passing through points \( B \) and \( C \) touches side \( AD \).

### Step 2: Finding \( r_1 \)
1. **Circle through \( A \) and \( B \)**: The center \( O \) of this circle lies directly above the midpoint of \( AB \) (which is \( A + \frac{a}{2} \) along the x-axis) and at a distance \( r_1 \) from points \( A \) and \( B \).
2. The vertical distance from the center \( O \) to side \( CD \) is \( r_1 - b \) (since the radius extends above the rectangle).
3. Using the Pythagorean theorem in triangle \( OAB \):
\[ (r_1 - b)^2 + \left(\frac{a}{2}\right)^2 = r_1^2 \]
Expanding this gives:
\[ r_1^2 - 2br_1 + b^2 + \frac{a^2}{4} = r_1^2 \]
Simplifying leads to:
\[ 2br_1 = b^2 + \frac{a^2}{4} \]
Thus, we find:
\[ r_1 = \frac{b^2 + \frac{a^2}{4}}{2b} = \frac{b}{2} + \frac{a^2}{8b} \]

### Step 3: Finding \( r_2 \)
1. **Circle through \( B \) and \( C \)**: The center \( P \) of this circle lies directly to the left of the midpoint of \( BC \) (which is \( B + \frac{b}{2} \) along the y-axis) and at a distance \( r_2 \) from points \( B \) and \( C \).
2. The horizontal distance from the center \( P \) to side \( AD \) is \( r_2 - a \).
3. Using the Pythagorean theorem in triangle \( PBC \):
\[ (r_2 - a)^2 + \left(\frac{b}{2}\right)^2 = r_2^2 \]
Expanding this gives:
\[ r_2^2 - 2ar_2 + a^2 + \frac{b^2}{4} = r_2^2 \]
Simplifying leads to:
\[ 2ar_2 = a^2 + \frac{b^2}{4} \]
Thus, we find:
\[ r_2 = \frac{a^2 + \frac{b^2}{4}}{2a} = \frac{a}{2} + \frac{b^2}{8a} \]

### Step 4: Finding \( r_1 + r_2 \)
Now we sum \( r_1 \) and \( r_2 \):
\[ r_1 + r_2 = \left( \frac{b}{2} + \frac{a^2}{8b} \right) + \left( \frac{a}{2} + \frac{b^2}{8a} \right) \]
Combining these gives:
\[ r_1 + r_2 = \frac{a + b}{2} + \left( \frac{a^2}{8b} + \frac{b^2}{8a} \right) \]

### Step 5: Finding the Minimum Value
To minimize \( r_1 + r_2 \), we can use the method of Lagrange multipliers or AM-GM inequality. By applying AM-GM:
\[ \frac{a^2}{8b} + \frac{b^2}{8a} \geq \frac{a + b}{4} \]
Thus:
\[ r_1 + r_2 \geq \frac{a + b}{2} + \frac{a + b}{4} = \frac{3(a + b)}{4} \]
The minimum value occurs when \( a = b \).

### Final Result
The minimum value of \( r_1 + r_2 \) is:
\[ \frac{(a + b)^2}{4ab} \]