Find the equation of the circle which cuts the circles `x^2+y^2-4x-6y+11=0` and `x^2+y^2-10x-4y+21=0` orthogonally and has the diameter along the line 2x+3y=7.

Answer

To find the equation of the circle that cuts the given circles orthogonally and has its diameter along the line \(2x + 3y = 7\), we can follow these steps:

### Step 1: Identify the given circles
The equations of the given circles are:
1. \(x^2 + y^2 - 4x - 6y + 11 = 0\)
2. \(x^2 + y^2 - 10x - 4y + 21 = 0\)

### Step 2: Rewrite the equations in standard form
We can rewrite the equations of the circles in standard form \( (x - h)^2 + (y - k)^2 = r^2 \).

1. For the first circle:
\[ x^2 + y^2 - 4x - 6y + 11 = 0 \implies (x - 2)^2 + (y - 3)^2 = 2 \]
Here, the center \(C_1\) is \((2, 3)\) and the radius \(r_1 = \sqrt{2}\).

2. For the second circle:
\[ x^2 + y^2 - 10x - 4y + 21 = 0 \implies (x - 5)^2 + (y - 2)^2 = 2 \]
Here, the center \(C_2\) is \((5, 2)\) and the radius \(r_2 = \sqrt{2}\).

### Step 3: Find the center of the required circle
The center of the required circle, \((h, k)\), lies on the line \(2x + 3y = 7\). Therefore, we have:
\[ 2h + 3k = 7 \quad \text{(1)} \]

### Step 4: Use the orthogonality condition
For two circles to cut orthogonally, the following condition must hold:
\[ r_1^2 + r_2^2 = d^2 \]
where \(d\) is the distance between the centers of the circles.

Calculating the distance \(d\) between \(C_1(2, 3)\) and \(C_2(5, 2)\):
\[ d = \sqrt{(5 - 2)^2 + (2 - 3)^2} = \sqrt{3^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10} \]

Now, substituting the values:
\[ r_1^2 + r_2^2 = 2 + 2 = 4 \]
Thus, we have:
\[ 4 = d^2 = (h - 2)^2 + (k - 3)^2 \]
Substituting \(d^2\):
\[ (h - 2)^2 + (k - 3)^2 = 4 \quad \text{(2)} \]

### Step 5: Substitute \(k\) from equation (1) into equation (2)
From equation (1):
\[ k = \frac{7 - 2h}{3} \]
Substituting \(k\) into equation (2):
\[ (h - 2)^2 + \left(\frac{7 - 2h}{3} - 3\right)^2 = 4 \]
Simplifying:
\[ \left(\frac{7 - 2h - 9}{3}\right)^2 = \left(\frac{-2 - 2h}{3}\right)^2 = \frac{(2 + 2h)^2}{9} \]
Thus, we have:
\[ (h - 2)^2 + \frac{(2 + 2h)^2}{9} = 4 \]

### Step 6: Solve for \(h\)
Multiply through by 9 to eliminate the fraction:
\[ 9(h - 2)^2 + (2 + 2h)^2 = 36 \]
Expanding:
\[ 9(h^2 - 4h + 4) + (4 + 8h + 4h^2) = 36 \]
Combining like terms:
\[ 9h^2 - 36h + 36 + 4h^2 + 8h + 4 = 36 \]
\[ 13h^2 - 28h + 4 = 0 \]
Using the quadratic formula:
\[ h = \frac{-(-28) \pm \sqrt{(-28)^2 - 4 \cdot 13 \cdot 4}}{2 \cdot 13} \]
\[ h = \frac{28 \pm \sqrt{784 - 208}}{26} = \frac{28 \pm \sqrt{576}}{26} = \frac{28 \pm 24}{26} \]
Calculating the two possible values for \(h\):
1. \(h = \frac{52}{26} = 2\)
2. \(h = \frac{4}{26} = \frac{2}{13}\)

### Step 7: Find corresponding \(k\) values
Using \(h = 2\):
\[ k = \frac{7 - 2(2)}{3} = \frac{3}{3} = 1 \]

Using \(h = \frac{2}{13}\):
\[ k = \frac{7 - 2(\frac{2}{13})}{3} = \frac{7 - \frac{4}{13}}{3} = \frac{\frac{91 - 4}{13}}{3} = \frac{87}{39} = \frac{29}{13} \]

### Step 8: Find \(r_1^2\)
Using the orthogonality condition again, we can find \(r_1^2\):
\[ r_1^2 = 4 - r_2^2 = 4 - 2 = 2 \]

### Step 9: Write the equation of the required circle
For \(h = 2\) and \(k = 1\):
\[ (x - 2)^2 + (y - 1)^2 = 2 \]
Expanding:
\[ x^2 - 4x + 4 + y^2 - 2y + 1 = 2 \]
\[ x^2 + y^2 - 4x - 2y + 3 = 0 \]

### Final Answer
The equation of the required circle is:
\[ x^2 + y^2 - 4x - 2y + 3 = 0 \]