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

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `90^(@)`

Answer

To find the angle subtended at the origin by the common chord of the circles given by the equations \(x^2 + y^2 - 4x - 4y = 0\) and \(x^2 + y^2 - 16 = 0\), we can follow these steps:

### Step 1: Rewrite the equations of the circles in standard form
1. **First Circle**:
\[ x^2 + y^2 - 4x - 4y = 0 \]
Completing the square:
\[ (x^2 - 4x) + (y^2 - 4y) = 0 \]
\[ (x - 2)^2 - 4 + (y - 2)^2 - 4 = 0 \]
\[ (x - 2)^2 + (y - 2)^2 = 8 \]
This represents a circle with center \((2, 2)\) and radius \(2\sqrt{2}\).

2. **Second Circle**:
\[ x^2 + y^2 - 16 = 0 \]
This can be rewritten as:
\[ x^2 + y^2 = 16 \]
This represents a circle with center \((0, 0)\) and radius \(4\).

### Step 2: Find the equation of the common chord
To find the common chord, we subtract the equations of the two circles:
\[ S_1 - S_2 = 0 \]
Where \(S_1\) is the equation of the first circle and \(S_2\) is the equation of the second circle:
\[ (x^2 + y^2 - 4x - 4y) - (x^2 + y^2 - 16) = 0 \]
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 \(-1\).

### Step 4: Calculate the angle with respect to the axes
The angle \(\theta\) that this line makes with the positive x-axis can be found using the tangent of the angle:
\[ \tan(\theta) = \text{slope} = -1 \]
This means:
\[ \theta = 135^\circ \quad \text{(since the slope is negative, it is in the second quadrant)} \]

### Step 5: Find the angle subtended at the origin
The angle subtended at the origin by the line \(x + y = 4\) with respect to the x-axis is \(135^\circ\). However, we need to find the angle between the two lines forming the chord. Since the line intersects both axes at equal distances, it subtends an angle of \(90^\circ\) at the origin.

### Conclusion
The angle subtended at the origin by the common chord of the two circles is \(90^\circ\).