The number of circles belonging to the system of circles `2(x^(2)+y^(2))+lambda x-(1+lambda^(2))y-10=0` and orthogonal to `x^(2)+y^(2)+4x+6y+3=0`, is

Options

A. 2
B. 1
C. 0
D. none of these

Answer

To solve the problem, we need to determine the number of circles from the system defined by the equation \(2(x^2 + y^2) + \lambda x - (1 + \lambda^2)y - 10 = 0\) that are orthogonal to the circle given by the equation \(x^2 + y^2 + 4x + 6y + 3 = 0\).

### Step 1: Convert the equations to standard form

1. **First Circle**:
\[ 2(x^2 + y^2) + \lambda x - (1 + \lambda^2)y - 10 = 0 \]
Divide the entire equation by 2:
\[ x^2 + y^2 + \frac{\lambda}{2} x - \frac{1 + \lambda^2}{2} y - 5 = 0 \]
Here, we can identify:
- \(g_1 = \frac{\lambda}{4}\)
- \(f_1 = -\frac{1 + \lambda^2}{4}\)
- \(c_1 = -5\)

2. **Second Circle**:
\[ x^2 + y^2 + 4x + 6y + 3 = 0 \]
Comparing with the standard form:
- \(g_2 = 2\)
- \(f_2 = 3\)
- \(c_2 = 3\)

### Step 2: Use the orthogonality condition

The condition for two circles to be orthogonal is given by:
\[ 2g_1g_2 + 2f_1f_2 = c_1 + c_2 \]
Substituting the values we found:
\[ 2\left(\frac{\lambda}{4}\right)(2) + 2\left(-\frac{1 + \lambda^2}{4}\right)(3) = -5 + 3 \]
This simplifies to:
\[ \frac{\lambda}{2} - \frac{3(1 + \lambda^2)}{2} = -2 \]
Multiplying through by 2 to eliminate the fraction:
\[ \lambda - 3(1 + \lambda^2) = -4 \]
Rearranging gives:
\[ -3\lambda^2 - \lambda + 1 = 0 \]

### Step 3: Solve the quadratic equation

The quadratic equation is:
\[ 3\lambda^2 + \lambda - 1 = 0 \]
Using the discriminant to find the number of solutions:
\[ D = b^2 - 4ac = 1^2 - 4 \cdot 3 \cdot (-1) = 1 + 12 = 13 \]
Since the discriminant \(D > 0\), there are two distinct real roots for \(\lambda\).

### Conclusion

Thus, there are **two circles** from the system that are orthogonal to the given circle.

### Final Answer:
The number of circles belonging to the system that are orthogonal to the given circle is **2**.

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