If a straight line through `C(-sqrt(8),sqrt(8))` making an angle of `135^ @` with the x-axis cuts the circle `x=5cos theta, y=5sin theta` in points A and B, then the length of AB is

Options

A. 6
B. 8
C. 10
D. none

Answer

To find the length of the line segment AB, where points A and B are the intersections of a line through point C(-√8, √8) making an angle of 135° with the x-axis and the circle defined by the parametric equations x = 5cos(θ) and y = 5sin(θ), we can follow these steps:

### Step 1: Determine the slope of the line
The slope \( m \) of a line making an angle \( \theta \) with the x-axis is given by:
\[ m = \tan(\theta) \]
For \( \theta = 135^\circ \):
\[ m = \tan(135^\circ) = -1 \]

### Step 2: Write the equation of the line
Using the point-slope form of the line equation, \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) = (-\sqrt{8}, \sqrt{8}) \):
\[ y - \sqrt{8} = -1(x + \sqrt{8}) \]
Rearranging gives:
\[ y - \sqrt{8} = -x - \sqrt{8} \]
\[ x + y = 0 \]

### Step 3: Identify the equation of the circle
The circle is defined by the parametric equations:
\[ x = 5\cos(\theta), \quad y = 5\sin(\theta) \]
The standard form of the circle is:
\[ x^2 + y^2 = r^2 \]
Here, \( r = 5 \), so:
\[ x^2 + y^2 = 25 \]

### Step 4: Find the intersection points of the line and the circle
Substituting the line equation \( y = -x \) into the circle equation:
\[ x^2 + (-x)^2 = 25 \]
\[ x^2 + x^2 = 25 \]
\[ 2x^2 = 25 \]
\[ x^2 = \frac{25}{2} \]
\[ x = \pm \frac{5}{\sqrt{2}} = \pm \frac{5\sqrt{2}}{2} \]

Now, substituting back to find \( y \):
For \( x = \frac{5\sqrt{2}}{2} \):
\[ y = -\frac{5\sqrt{2}}{2} \]
For \( x = -\frac{5\sqrt{2}}{2} \):
\[ y = \frac{5\sqrt{2}}{2} \]

Thus, the points of intersection A and B are:
\[ A\left(\frac{5\sqrt{2}}{2}, -\frac{5\sqrt{2}}{2}\right), \quad B\left(-\frac{5\sqrt{2}}{2}, \frac{5\sqrt{2}}{2}\right) \]

### Step 5: Calculate the length of AB
The length \( AB \) can be calculated using the distance formula:
\[ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates of points A and B:
\[ AB = \sqrt{\left(-\frac{5\sqrt{2}}{2} - \frac{5\sqrt{2}}{2}\right)^2 + \left(\frac{5\sqrt{2}}{2} - \left(-\frac{5\sqrt{2}}{2}\right)\right)^2} \]
\[ = \sqrt{\left(-5\sqrt{2}\right)^2 + \left(5\sqrt{2}\right)^2} \]
\[ = \sqrt{(25 \cdot 2) + (25 \cdot 2)} = \sqrt{50 + 50} = \sqrt{100} = 10 \]

### Final Answer
The length of segment AB is \( 10 \).

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