Two circles of radii `r_1 and r_2`, are both touching the coordinate axes and intersecting each other orthogonally. The value of `r_1/r_2` (where `r_1 > r_2`) equals -

Options

A. 2
B. `2+sqrt3`
C. `3+sqrt2`
D. 3

Answer

To solve the problem, we need to find the ratio \( \frac{r_1}{r_2} \) for two circles that touch the coordinate axes and intersect orthogonally. Here’s a step-by-step solution:

### Step 1: Understand the Position of the Circles
Both circles touch the coordinate axes. Therefore, the centers of the circles will be at the coordinates:
- For the smaller circle with radius \( r_2 \): Center at \( (r_2, r_2) \)
- For the larger circle with radius \( r_1 \): Center at \( (r_1, r_1) \)

### Step 2: Write the Equations of the Circles
The equation of a circle in standard form is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
where \( (h, k) \) is the center and \( r \) is the radius.

For the smaller circle:
\[ (x - r_2)^2 + (y - r_2)^2 = r_2^2 \]
Expanding this gives:
\[ x^2 - 2r_2x + r_2^2 + y^2 - 2r_2y + r_2^2 = r_2^2 \]
Simplifying, we have:
\[ x^2 + y^2 - 2r_2x - 2r_2y + r_2^2 = 0 \quad \text{(Equation 1)} \]

For the larger circle:
\[ (x - r_1)^2 + (y - r_1)^2 = r_1^2 \]
Expanding this gives:
\[ x^2 - 2r_1x + r_1^2 + y^2 - 2r_1y + r_1^2 = r_1^2 \]
Simplifying, we have:
\[ x^2 + y^2 - 2r_1x - 2r_1y + r_1^2 = 0 \quad \text{(Equation 2)} \]

### Step 3: Use the Condition of Orthogonality
The circles intersect orthogonally if:
\[ 2f_1f_2 + 2g_1g_2 = c_1 + c_2 \]
From the equations:
- For the smaller circle (Equation 1):
- \( g_1 = -r_2 \), \( f_1 = -r_2 \), \( c_1 = r_2^2 \)

- For the larger circle (Equation 2):
- \( g_2 = -r_1 \), \( f_2 = -r_1 \), \( c_2 = r_1^2 \)

Substituting these into the orthogonality condition:
\[ 2(-r_2)(-r_1) + 2(-r_2)(-r_1) = r_2^2 + r_1^2 \]
This simplifies to:
\[ 4r_1r_2 = r_1^2 + r_2^2 \]

### Step 4: Rearranging the Equation
Rearranging gives:
\[ r_1^2 - 4r_1r_2 + r_2^2 = 0 \]

### Step 5: Let \( x = \frac{r_1}{r_2} \)
Substituting \( r_1 = xr_2 \) into the equation:
\[ (xr_2)^2 - 4(xr_2)(r_2) + r_2^2 = 0 \]
This simplifies to:
\[ x^2r_2^2 - 4xr_2^2 + r_2^2 = 0 \]
Dividing through by \( r_2^2 \) (since \( r_2 \neq 0 \)):
\[ x^2 - 4x + 1 = 0 \]

### Step 6: Solve the Quadratic Equation
Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3} \]

### Step 7: Determine the Correct Ratio
Since \( r_1 > r_2 \), we take the positive value:
\[ \frac{r_1}{r_2} = 2 + \sqrt{3} \]

### Final Answer
Thus, the value of \( \frac{r_1}{r_2} \) is:
\[ \frac{r_1}{r_2} = 2 + \sqrt{3} \]

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