Two circles touching both the axes intersect at (3,-2) then the coordinates of their other point of intersection is

Options

A. (2,3)
B. (2,-3)
C. (-2,3)
D. (-2,-3)

Answer

To solve the problem of finding the coordinates of the other point of intersection of two circles that touch both axes and intersect at the point (3, -2), we can follow these steps:

### Step 1: Understand the Problem
We have two circles that touch both the x-axis and y-axis and intersect at the point (3, -2). We need to find the coordinates of their other point of intersection.

**Hint:** Visualize the circles and their properties. Since they touch both axes, their centers will be located along the line y = -x.

### Step 2: Identify the Line of Centers
The centers of the circles must lie on the line y = -x, which can also be expressed as x + y = 0.

**Hint:** This line represents the symmetry of the circles with respect to the origin.

### Step 3: Use the Reflection Property
To find the other point of intersection, we can reflect the point (3, -2) across the line y = -x. The reflection of a point (x1, y1) across the line ax + by + c = 0 can be calculated using the formula:
- \( h = x_1 - \frac{2a(ax_1 + by_1 + c)}{a^2 + b^2} \)
- \( k = y_1 - \frac{2b(ax_1 + by_1 + c)}{a^2 + b^2} \)

For our line x + y = 0, we have:
- \( a = 1, b = 1, c = 0 \)
- The point (x1, y1) = (3, -2)

**Hint:** Substitute the values into the reflection formula.

### Step 4: Calculate the Reflection
1. Calculate \( ax_1 + by_1 + c \):
- \( 1(3) + 1(-2) + 0 = 3 - 2 + 0 = 1 \)

2. Substitute into the formulas:
- \( h = 3 - \frac{2(1)(1)}{1^2 + 1^2} = 3 - \frac{2}{2} = 3 - 1 = 2 \)
- \( k = -2 - \frac{2(1)(1)}{1^2 + 1^2} = -2 - \frac{2}{2} = -2 - 1 = -3 \)

**Hint:** Ensure you perform the arithmetic correctly.

### Step 5: Conclusion
The coordinates of the other point of intersection of the two circles are (2, -3).

**Final Answer:** (2, -3)