The abscissa of two A and B are the roots of the equation `x ^(2) - 4x - 6=0` and their ordinates are the roots of `x ^(2) + 5x - 4=0.` If r is the radius of circle with AB as diameter, then r is equal to

Options

A. `9/2`
B. `9`
C. `7/2`
D. `7`

Answer

To solve the problem step by step, we will first find the roots of the given equations and then use them to calculate the radius of the circle with AB as the diameter.

### Step 1: Find the roots of the first equation
The first equation given is:
\[ x^2 - 4x - 6 = 0 \]

To find the roots, 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 (abscissas) are:
\[ x_1 = 2 + \sqrt{10}, \quad x_2 = 2 - \sqrt{10} \]

### Step 2: Find the roots of the second equation
The second equation is:
\[ x^2 + 5x - 4 = 0 \]

Using the quadratic formula again:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = 5, c = -4 \).

Calculating the discriminant:
\[ b^2 - 4ac = 5^2 - 4 \cdot 1 \cdot (-4) = 25 + 16 = 41 \]

Now substituting back into the quadratic formula:
\[ x = \frac{-5 \pm \sqrt{41}}{2} \]

Thus, the roots (ordinates) are:
\[ y_1 = \frac{-5 + \sqrt{41}}{2}, \quad y_2 = \frac{-5 - \sqrt{41}}{2} \]

### Step 3: Calculate the distance AB
The distance \( AB \) (which is the diameter of the circle) can be calculated using the distance formula:
\[ AB = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} \]

First, calculate \( x_1 - x_2 \):
\[ x_1 - x_2 = (2 + \sqrt{10}) - (2 - \sqrt{10}) = 2\sqrt{10} \]

Now calculate \( y_1 - y_2 \):
\[ y_1 - y_2 = \left(\frac{-5 + \sqrt{41}}{2}\right) - \left(\frac{-5 - \sqrt{41}}{2}\right) = \frac{2\sqrt{41}}{2} = \sqrt{41} \]

Now substitute these into the distance formula:
\[ AB = \sqrt{(2\sqrt{10})^2 + (\sqrt{41})^2} = \sqrt{4 \cdot 10 + 41} = \sqrt{40 + 41} = \sqrt{81} = 9 \]

### Step 4: Calculate the radius \( r \)
Since \( AB \) is the diameter of the circle:
\[ AB = 2r \]
Thus,
\[ 2r = 9 \]
So,
\[ r = \frac{9}{2} \]

### Final Answer
The radius \( r \) of the circle is:
\[ r = \frac{9}{2} \]

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