The circle with centres (2,3) and intersecting `x^2+y^2-4x+2y-7=0` orthogonally has the radius

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to find the radius of the circle that intersects orthogonally with the given circle. Let's break down the solution step by step.

### Step 1: Identify the given circle
The equation of the given circle is:
\[ x^2 + y^2 - 4x + 2y - 7 = 0 \]

We can rewrite this equation in standard form by completing the square.

### Step 2: Complete the square
1. Rearranging the terms:
\[ (x^2 - 4x) + (y^2 + 2y) = 7 \]

2. Completing the square for \(x\):
\[ x^2 - 4x = (x - 2)^2 - 4 \]

3. Completing the square for \(y\):
\[ y^2 + 2y = (y + 1)^2 - 1 \]

4. Substituting back into the equation:
\[ (x - 2)^2 - 4 + (y + 1)^2 - 1 = 7 \]
\[ (x - 2)^2 + (y + 1)^2 = 12 \]

Thus, the center of the given circle is \((2, -1)\) and its radius \(r_2 = \sqrt{12} = 2\sqrt{3}\).

### Step 3: Identify the center of the second circle
The center of the second circle is given as \((2, 3)\).

### Step 4: Calculate the distance between the centers
The distance \(d\) between the centers \((2, 3)\) and \((2, -1)\) is:
\[ d = \sqrt{(2 - 2)^2 + (3 - (-1))^2} = \sqrt{0 + 4^2} = 4 \]

### Step 5: Use the orthogonality condition
For two circles to intersect orthogonally, the following condition must hold:
\[ r_1^2 + r_2^2 = d^2 \]
Where \(r_1\) is the radius of the first circle and \(r_2\) is the radius of the second circle.

### Step 6: Substitute the known values
We know:
- \(r_2 = 2\sqrt{3}\)
- \(d = 4\)

Substituting into the orthogonality condition:
\[ r_1^2 + (2\sqrt{3})^2 = 4^2 \]
\[ r_1^2 + 12 = 16 \]

### Step 7: Solve for \(r_1^2\)
\[ r_1^2 = 16 - 12 \]
\[ r_1^2 = 4 \]

### Step 8: Find \(r_1\)
Taking the square root:
\[ r_1 = \sqrt{4} = 2 \]

### Conclusion
The radius of the circle that intersects orthogonally with the given circle is:
\[ \boxed{2} \]