Let roots of `x^2-4x-6=0` are the abscissa and roots of `y^2+2y-7=0` are the ordinates of the end of diameter of the circle `x^2+y^2+2ax+2by+c=0` then `a+b-c` is equal to

Options

A. 10
B. 11
C. 12
D. -12

Answer

To solve the problem, we need to find the values of \( a \), \( b \), and \( c \) from the equations given and then compute \( a + b - c \).

### Step 1: Find the roots of the first equation \( x^2 - 4x - 6 = 0 \).
We can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = -4, c = -6 \).

Calculating the discriminant:
\[ b^2 - 4ac = (-4)^2 - 4 \cdot 1 \cdot (-6) = 16 + 24 = 40 \]
Now, substituting back into the quadratic formula:
\[ x = \frac{4 \pm \sqrt{40}}{2} = \frac{4 \pm 2\sqrt{10}}{2} = 2 \pm \sqrt{10} \]
Thus, the roots are:
\[ x_1 = 2 + \sqrt{10}, \quad x_2 = 2 - \sqrt{10} \]

### Step 2: Find the roots of the second equation \( y^2 + 2y - 7 = 0 \).
Using the quadratic formula again:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = 2, c = -7 \).

Calculating the discriminant:
\[ b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot (-7) = 4 + 28 = 32 \]
Now substituting back into the quadratic formula:
\[ y = \frac{-2 \pm \sqrt{32}}{2} = \frac{-2 \pm 4\sqrt{2}}{2} = -1 \pm 2\sqrt{2} \]
Thus, the roots are:
\[ y_1 = -1 + 2\sqrt{2}, \quad y_2 = -1 - 2\sqrt{2} \]

### Step 3: Substitute the values into the circle equation.
The equation of the circle is given as:
\[ x^2 + y^2 + 2ax + 2by + c = 0 \]
From the roots we found, we can express \( x^2 + y^2 \) in terms of \( a \), \( b \), and \( c \).

Using the values of \( x_1 \) and \( x_2 \):
\[ x^2 = 4x + 6 \quad \text{(from the first equation)} \]
Using the values of \( y_1 \) and \( y_2 \):
\[ y^2 = 7 - 2y \quad \text{(from the second equation)} \]

### Step 4: Combine the equations.
Now we can write:
\[ x^2 + y^2 = (4x + 6) + (7 - 2y) = 4x - 2y + 13 \]
Substituting this into the circle equation:
\[ 4x - 2y + 13 + 2ax + 2by + c = 0 \]

### Step 5: Collect coefficients.
Rearranging gives:
\[ (4 + 2a)x + (-2 + 2b)y + (c + 13) = 0 \]
For this to hold for all \( x \) and \( y \), the coefficients must equal zero:
1. \( 4 + 2a = 0 \)
2. \( -2 + 2b = 0 \)
3. \( c + 13 = 0 \)

### Step 6: Solve for \( a \), \( b \), and \( c \).
From \( 4 + 2a = 0 \):
\[ 2a = -4 \implies a = -2 \]
From \( -2 + 2b = 0 \):
\[ 2b = 2 \implies b = 1 \]
From \( c + 13 = 0 \):
\[ c = -13 \]

### Step 7: Calculate \( a + b - c \).
Now, substituting the values:
\[ a + b - c = -2 + 1 - (-13) = -2 + 1 + 13 = 12 \]

Thus, the final answer is:
\[ \boxed{12} \]