Let circle cuts ortholognally each of the three circles `x^(2) + y^(2) + 3x + 4y + 11 = 0, x^(2) + y^(2) - 3x + 7y - 1 = 0` and `x^(2) + y^(2) + 2x = 0`

Options

A. The centre of the circle is `(-3, -2)`
B. Radius of the circle is 3
C. Equation of chord of contact of tangents drawn from `(2,4)` is `5x + 6y - 18 = 0`
D. Length of tangent from (2, 4) to the circle is `sqrt(3)`

Answer

To find the circle that cuts orthogonally with the given three circles, we will follow these steps:

### Step 1: Write the equations of the given circles
The equations of the three circles are:
1. \( C_1: x^2 + y^2 + 3x + 4y + 11 = 0 \)
2. \( C_2: x^2 + y^2 - 3x + 7y - 1 = 0 \)
3. \( C_3: x^2 + y^2 + 2x = 0 \)

### Step 2: Convert the equations to standard form
To find the center and radius of each circle, we can rewrite the equations in standard form.

**For \( C_1 \):**
\[ x^2 + 3x + y^2 + 4y + 11 = 0 \]
Completing the square:
\[ (x + \frac{3}{2})^2 - \frac{9}{4} + (y + 2)^2 - 4 + 11 = 0 \]
\[ (x + \frac{3}{2})^2 + (y + 2)^2 = \frac{9}{4} - 4 + 11 = \frac{9}{4} - \frac{16}{4} + \frac{44}{4} = \frac{37}{4} \]
Thus, the center is \( (-\frac{3}{2}, -2) \) and radius \( r_1 = \sqrt{\frac{37}{4}} = \frac{\sqrt{37}}{2} \).

**For \( C_2 \):**
\[ x^2 - 3x + y^2 + 7y - 1 = 0 \]
Completing the square:
\[ (x - \frac{3}{2})^2 - \frac{9}{4} + (y + \frac{7}{2})^2 - \frac{49}{4} - 1 = 0 \]
\[ (x - \frac{3}{2})^2 + (y + \frac{7}{2})^2 = \frac{9}{4} + \frac{49}{4} + 1 = \frac{9 + 49 + 4}{4} = \frac{62}{4} = \frac{31}{2} \]
Thus, the center is \( (\frac{3}{2}, -\frac{7}{2}) \) and radius \( r_2 = \sqrt{\frac{31}{2}} \).

**For \( C_3 \):**
\[ x^2 + 2x + y^2 = 0 \]
Completing the square:
\[ (x + 1)^2 - 1 + y^2 = 0 \]
\[ (x + 1)^2 + y^2 = 1 \]
Thus, the center is \( (-1, 0) \) and radius \( r_3 = 1 \).

### Step 3: Find the radical center
The radical center of the three circles can be found by solving the equations of the radical axes.

**Radical axis of \( C_1 \) and \( C_2 \):**
\[ C_1 - C_2 = 0 \Rightarrow (x^2 + y^2 + 3x + 4y + 11) - (x^2 + y^2 - 3x + 7y - 1) = 0 \]
This simplifies to:
\[ 6x - 3y + 12 = 0 \Rightarrow 2x - y + 4 = 0 \quad \text{(Equation 1)} \]

**Radical axis of \( C_2 \) and \( C_3 \):**
\[ C_2 - C_3 = 0 \Rightarrow (x^2 + y^2 - 3x + 7y - 1) - (x^2 + 2x + y^2) = 0 \]
This simplifies to:
\[ -5x + 7y - 1 = 0 \quad \text{(Equation 2)} \]

### Step 4: Solve the system of equations
Now we solve the two equations:
1. \( 2x - y + 4 = 0 \)
2. \( -5x + 7y - 1 = 0 \)

From Equation 1, we can express \( y \):
\[ y = 2x + 4 \]

Substituting \( y \) in Equation 2:
\[ -5x + 7(2x + 4) - 1 = 0 \]
\[ -5x + 14x + 28 - 1 = 0 \Rightarrow 9x + 27 = 0 \Rightarrow x = -3 \]

Substituting \( x = -3 \) back into Equation 1:
\[ y = 2(-3) + 4 = -6 + 4 = -2 \]

Thus, the radical center is \( (-3, -2) \).

### Step 5: Find the radius
To find the radius of the circle that cuts orthogonally, we calculate the length of the tangent from the radical center to any of the circles, say \( C_1 \):
\[ \text{Length of tangent} = \sqrt{C_1(-3, -2)} \]
Substituting into \( C_1 \):
\[ (-3)^2 + (-2)^2 + 3(-3) + 4(-2) + 11 = 9 + 4 - 9 - 8 + 11 = 7 \]
Thus, the radius \( r = \sqrt{7} \).

### Final Answer
The circle that cuts orthogonally with the three given circles has its center at \( (-3, -2) \) and radius \( \sqrt{7} \).

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