If the circles `x ^(2) + y ^(2) + 5x -6y-1=0 and x ^(2) + y^(2) +ax -y +1=0` intersect orthogonally (the tangents at the point of intersection of the circles are at right angles), the value of a is

Options

A. `6//5`
B. `5//6`
C. `-6//5`
D. `-5//6`

Answer

To solve the problem, we need to find the value of \( a \) such that the circles given by the equations \( x^2 + y^2 + 5x - 6y - 1 = 0 \) and \( x^2 + y^2 + ax - y + 1 = 0 \) intersect orthogonally.

### Step 1: Identify the coefficients from the equations of the circles

The general form of a circle is given by:
\[ x^2 + y^2 + 2g x + 2f y + c = 0 \]

For the first circle \( x^2 + y^2 + 5x - 6y - 1 = 0 \):
- \( 2g_1 = 5 \) → \( g_1 = \frac{5}{2} \)
- \( 2f_1 = -6 \) → \( f_1 = -3 \)
- \( c_1 = -1 \)

For the second circle \( x^2 + y^2 + ax - y + 1 = 0 \):
- \( 2g_2 = a \) → \( g_2 = \frac{a}{2} \)
- \( 2f_2 = -1 \) → \( f_2 = -\frac{1}{2} \)
- \( c_2 = 1 \)

### Step 2: Use the condition for orthogonality of circles

The circles intersect orthogonally if the following condition holds:
\[ 2g_1g_2 + 2f_1f_2 = c_1 + c_2 \]

Substituting the values we found:
\[ 2 \left( \frac{5}{2} \right) \left( \frac{a}{2} \right) + 2(-3)\left(-\frac{1}{2}\right) = -1 + 1 \]

### Step 3: Simplify the equation

Calculating each term:
- The left side becomes:
\[ 2 \cdot \frac{5}{2} \cdot \frac{a}{2} = \frac{5a}{2} \]
\[ 2 \cdot (-3) \cdot \left(-\frac{1}{2}\right) = 3 \]
So we have:
\[ \frac{5a}{2} + 3 = 0 \]

### Step 4: Solve for \( a \)

Now, we can solve for \( a \):
\[ \frac{5a}{2} = -3 \]
Multiplying both sides by 2:
\[ 5a = -6 \]
Dividing by 5:
\[ a = -\frac{6}{5} \]

### Final Answer

The value of \( a \) is:
\[ \boxed{-\frac{6}{5}} \]