The common chord of `x^2 + y^2 - 4x - 4y = 0` and `x^2 + y^2 = 16` substends at the origin an angle equal to

Options

A. `pi/6`
B. `pi/4`
C. `pi/3`
D. `pi/2`

Answer

To find the angle that the common chord of the circles \(x^2 + y^2 - 4x - 4y = 0\) and \(x^2 + y^2 = 16\) subtends at the origin, we will follow these steps:

### Step 1: Rewrite the equations of the circles
The first equation can be rewritten by completing the square:
\[ x^2 - 4x + y^2 - 4y = 0 \]
Completing the square for \(x\) and \(y\):
\[ (x - 2)^2 - 4 + (y - 2)^2 - 4 = 0 \]
This simplifies to:
\[ (x - 2)^2 + (y - 2)^2 = 8 \]
This represents a circle centered at \((2, 2)\) with a radius of \(2\sqrt{2}\).

The second equation is already in standard form:
\[ x^2 + y^2 = 16 \]
This represents a circle centered at \((0, 0)\) with a radius of \(4\).

### Step 2: Find the equation of the common chord
The common chord can be found using the equation \(S_1 - S_2 = 0\), where \(S_1\) and \(S_2\) are the equations of the circles.

From the first circle:
\[ S_1: x^2 + y^2 - 4x - 4y = 0 \]
From the second circle:
\[ S_2: x^2 + y^2 - 16 = 0 \]
Now, we subtract \(S_2\) from \(S_1\):
\[ S_1 - S_2 = (x^2 + y^2 - 4x - 4y) - (x^2 + y^2 - 16) = 0 \]
This simplifies to:
\[ -4x - 4y + 16 = 0 \]
Rearranging gives:
\[ 4x + 4y = 16 \quad \Rightarrow \quad x + y = 4 \]

### Step 3: Determine the angle subtended at the origin
The line \(x + y = 4\) can be rewritten in slope-intercept form:
\[ y = -x + 4 \]
The slope of this line is \(-1\).

To find the angle \(\theta\) that this line makes with the positive x-axis, we use the formula:
\[ \tan(\theta) = \text{slope} = -1 \]
Thus, the angle \(\theta\) is:
\[ \theta = \tan^{-1}(-1) = \frac{3\pi}{4} \text{ or } \frac{7\pi}{4} \]
However, since we are interested in the angle subtended at the origin, we consider the angle between the two lines that the chord represents.

### Step 4: Find the angle subtended
The angle subtended by the line \(x + y = 4\) at the origin is:
\[ \theta = \frac{\pi}{2} \]

### Conclusion
The angle subtended at the origin by the common chord is \(\frac{\pi}{2}\).