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 thorugh `B` & `C` and touching side `AD`. Let `r_(1)` & `r_(2)` be the radii of these two circle respectively.
`(r_(1))/(r_(2))` equals

Options

A. `a/b((4b^(2)+a^(2))/(4a^(2)+b^(2)))`
B. `b/a ((4a^(2)+b(2))/(4b^(2)+a^(2)))`
C. `a/b((4b^(2)-a^(2))/(4a^(2)-b^(2)))`
D. `b/a((a^(2)-4b^(2))/(4a^(2)-b^(2)))`

Explanation

To solve the problem, we need to find the ratio of the radii \( r_1 \) and \( r_2 \) of the two circles drawn in rectangle \( ABCD \).

### Step 1: Define the Geometry
Let \( AB = a \) and \( BC = b \). The first circle passes through points \( A \) and \( B \) and touches side \( CD \). The second circle passes through points \( B \) and \( C \) and touches side \( AD \).

### Step 2: Analyze the First Circle
Let the center of the first circle be \( O_1 \) and the radius be \( r_1 \). The distance from \( O_1 \) to line \( CD \) is \( r_1 \). The distance from \( O_1 \) to point \( B \) is \( r_1 \) as well.

In triangle \( O_1MB \) (where \( M \) is the midpoint of \( AB \)):
- \( MB = \frac{a}{2} \)
- The height from \( O_1 \) to line \( CD \) is \( r_1 - b \).

Using the Pythagorean theorem:
\[ r_1^2 = (r_1 - b)^2 + \left(\frac{a}{2}\right)^2 \]

### Step 3: Expand and Simplify
Expanding the equation:
\[ r_1^2 = (r_1^2 - 2br_1 + b^2) + \frac{a^2}{4} \]
Cancelling \( r_1^2 \) from both sides:
\[ 0 = -2br_1 + b^2 + \frac{a^2}{4} \]
Rearranging gives:
\[ 2br_1 = b^2 + \frac{a^2}{4} \]
Thus,
\[ r_1 = \frac{b^2 + \frac{a^2}{4}}{2b} = \frac{4b^2 + a^2}{8b} \]

### Step 4: Analyze the Second Circle
Let the center of the second circle be \( O_2 \) and the radius be \( r_2 \). The distance from \( O_2 \) to line \( AD \) is \( r_2 \). The distance from \( O_2 \) to point \( C \) is \( r_2 \) as well.

In triangle \( O_2CN \) (where \( N \) is the midpoint of \( BC \)):
- \( CN = \frac{b}{2} \)
- The height from \( O_2 \) to line \( AD \) is \( A - r_2 \).

Using the Pythagorean theorem:
\[ r_2^2 = (A - r_2)^2 + \left(\frac{b}{2}\right)^2 \]

### Step 5: Expand and Simplify
Expanding the equation:
\[ r_2^2 = (A^2 - 2Ar_2 + r_2^2) + \frac{b^2}{4} \]
Cancelling \( r_2^2 \) from both sides:
\[ 0 = -2Ar_2 + A^2 + \frac{b^2}{4} \]
Rearranging gives:
\[ 2Ar_2 = A^2 + \frac{b^2}{4} \]
Thus,
\[ r_2 = \frac{A^2 + \frac{b^2}{4}}{2A} = \frac{4A^2 + b^2}{8A} \]

### Step 6: Find the Ratio \( \frac{r_1}{r_2} \)
Now, we can find the ratio:
\[ \frac{r_1}{r_2} = \frac{\frac{4b^2 + a^2}{8b}}{\frac{4a^2 + b^2}{8a}} = \frac{(4b^2 + a^2) \cdot a}{(4a^2 + b^2) \cdot b} \]

### Final Result
Thus, the ratio \( \frac{r_1}{r_2} \) is:
\[ \frac{r_1}{r_2} = \frac{a(4b^2 + a^2)}{b(4a^2 + b^2)} \]