If `x^2+y^2+2ax+2by+c=0` is a circle whose diameter is PQ such that abscissa of P,Q are roots of equation `x^2-4x-6=0` and ordinates are roots of equation `y^2+2y-7=0` , then value of `(a+b-c) is

Options

A. 2
B. 10
C. 8
D. 6

Answer

To solve the problem, we need to analyze the given equations and extract the necessary information to find the values of \(a\), \(b\), and \(c\).

### Step 1: Find the roots of the equations for the abscissa and ordinate.
The abscissa of points \(P\) and \(Q\) are the roots of the equation:
\[ x^2 - 4x - 6 = 0 \]
To find the roots, we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 1\), \(b = -4\), and \(c = -6\).

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

### Step 2: Find the roots of the equation for the ordinates.
The ordinates of points \(P\) and \(Q\) are the roots of the equation:
\[ y^2 + 2y - 7 = 0 \]
Again, using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 1\), \(b = 2\), and \(c = -7\).

Calculating the discriminant:
\[ b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot (-7) = 4 + 28 = 32 \]
Now applying 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 (ordinates) are:
\[ y_1 = -1 + 2\sqrt{2}, \quad y_2 = -1 - 2\sqrt{2} \]

### Step 3: Write the equation of the circle.
The equation of a circle with diameter endpoints \(P(x_1, y_1)\) and \(Q(x_2, y_2)\) can be expressed as:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
where \(h\) and \(k\) are the midpoints of \(P\) and \(Q\), and \(r\) is the radius.

The center \((h, k)\) is:
\[ h = \frac{x_1 + x_2}{2} = \frac{(2 + \sqrt{10}) + (2 - \sqrt{10})}{2} = \frac{4}{2} = 2 \]
\[ k = \frac{y_1 + y_2}{2} = \frac{(-1 + 2\sqrt{2}) + (-1 - 2\sqrt{2})}{2} = \frac{-2}{2} = -1 \]

The radius \(r\) can be calculated as:
\[ r = \frac{\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}}{2} \]
Calculating \(x_1 - x_2\) and \(y_1 - y_2\):
\[ x_1 - x_2 = (2 + \sqrt{10}) - (2 - \sqrt{10}) = 2\sqrt{10} \]
\[ y_1 - y_2 = (-1 + 2\sqrt{2}) - (-1 - 2\sqrt{2}) = 4\sqrt{2} \]
Thus,
\[ r = \frac{\sqrt{(2\sqrt{10})^2 + (4\sqrt{2})^2}}{2} = \frac{\sqrt{40 + 32}}{2} = \frac{\sqrt{72}}{2} = \frac{6\sqrt{2}}{2} = 3\sqrt{2} \]

### Step 4: Write the standard form of the circle's equation.
The standard form of the circle's equation is:
\[ (x - 2)^2 + (y + 1)^2 = (3\sqrt{2})^2 \]
Expanding this gives:
\[ (x^2 - 4x + 4) + (y^2 + 2y + 1) = 18 \]
Combining terms:
\[ x^2 + y^2 - 4x + 2y - 13 = 0 \]

### Step 5: Compare with the general form.
The general form of the circle's equation is:
\[ x^2 + y^2 + 2ax + 2by + c = 0 \]
Comparing coefficients, we have:
\[ 2a = -4 \implies a = -2 \]
\[ 2b = 2 \implies b = 1 \]
\[ c = -13 \]

### Step 6: Calculate \(a + b - c\).
Now we can find \(a + b - c\):
\[ a + b - c = -2 + 1 - (-13) = -2 + 1 + 13 = 12 \]

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