Two circles `x^(2) + y^(2) + ax + ay - 7 = 0` and `x^(2) + y^(2) - 10x + 2ay + 1 = 0` will cut orthogonally if the value of a is

Options

A. 3
B. 2
C. `-2`
D. `-3`

Answer

To determine the value of \( a \) for which the two circles cut orthogonally, we can follow these steps:

### Step 1: Identify the equations of the circles
The equations of the circles are given as:
1. Circle 1: \( x^2 + y^2 + ax + ay - 7 = 0 \)
2. Circle 2: \( x^2 + y^2 - 10x + 2ay + 1 = 0 \)

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

For Circle 1:
- Coefficients: \( g_1 = \frac{a}{2}, f_1 = \frac{a}{2}, c_1 = -7 \)

For Circle 2:
- Coefficients: \( g_2 = -5, f_2 = a, c_2 = 1 \)

### Step 3: Use the orthogonality condition
Two circles cut orthogonally if the following condition holds:
\[ g_1 g_2 + f_1 f_2 = c_1 + c_2 \]

Substituting the values:
\[ \left(\frac{a}{2}\right)(-5) + \left(\frac{a}{2}\right)(a) = -7 + 1 \]
This simplifies to:
\[ -\frac{5a}{2} + \frac{a^2}{2} = -6 \]

### Step 4: Clear the fractions
To eliminate the fractions, multiply the entire equation by 2:
\[ -a^2 + 5a = -12 \]
Rearranging gives:
\[ a^2 - 5a - 12 = 0 \]

### Step 5: Factor the quadratic equation
Now, we can factor the quadratic equation:
\[ (a - 6)(a + 2) = 0 \]

### Step 6: Solve for \( a \)
Setting each factor to zero gives:
1. \( a - 6 = 0 \) → \( a = 6 \)
2. \( a + 2 = 0 \) → \( a = -2 \)

Thus, the values of \( a \) for which the circles cut orthogonally are \( a = 6 \) and \( a = -2 \).

### Final Answer:
The values of \( a \) are \( a = 6 \) and \( a = -2 \).

---