The equation of the line joining origin to the points of intersection of the curve `x^2+y^2=a^2` and `x^2+y^2-ax-ay=0`, is

Options

A. `x^2-y^2=0`
B. `xy=0`
C. `xy-x^2=0`
D. `y^2+xy=0`

Explanation

To find the equation of the line joining the origin to the points of intersection of the curves \(x^2 + y^2 = a^2\) and \(x^2 + y^2 - ax - ay = 0\), we can follow these steps:

### Step 1: Understand the curves
The first equation, \(x^2 + y^2 = a^2\), represents a circle centered at the origin with radius \(a\).

The second equation, \(x^2 + y^2 - ax - ay = 0\), can be rearranged to identify it as a pair of straight lines.

### Step 2: Rearranging the second equation
Rearranging the second equation gives:
\[ x^2 + y^2 = ax + ay \]
This can be rewritten as:
\[ x^2 - ax + y^2 - ay = 0 \]

### Step 3: Completing the square
To express this in a more useful form, we complete the square for both \(x\) and \(y\):
\[ (x - \frac{a}{2})^2 + (y - \frac{a}{2})^2 = \frac{a^2}{2} \]
This represents a circle centered at \((\frac{a}{2}, \frac{a}{2})\) with radius \(\frac{a}{\sqrt{2}}\).

### Step 4: Finding points of intersection
To find the points of intersection of the two curves, we can substitute \(x^2 + y^2 = a^2\) into the rearranged second equation:
\[ a^2 = ax + ay \]
This simplifies to:
\[ ax + ay = a^2 \]
Dividing through by \(a\) (assuming \(a \neq 0\)):
\[ x + y = a \]

### Step 5: Parametric representation
From \(x + y = a\), we can express \(y\) in terms of \(x\):
\[ y = a - x \]
Substituting this into the circle equation:
\[ x^2 + (a - x)^2 = a^2 \]
Expanding gives:
\[ x^2 + (a^2 - 2ax + x^2) = a^2 \]
Combining like terms:
\[ 2x^2 - 2ax = 0 \]
Factoring out \(2x\):
\[ 2x(x - a) = 0 \]
Thus, \(x = 0\) or \(x = a\).

### Step 6: Finding corresponding \(y\) values
If \(x = 0\), then \(y = a\).
If \(x = a\), then \(y = 0\).
The points of intersection are \((0, a)\) and \((a, 0)\).

### Step 7: Equation of the line from the origin
The line joining the origin \((0, 0)\) to the point \((0, a)\) is the vertical line \(x = 0\).
The line joining the origin to the point \((a, 0)\) is the horizontal line \(y = 0\).

### Final Step: General equation of the line
The equation of the line joining the origin to the points of intersection can be expressed in slope-intercept form. The line through \((0, a)\) is vertical, and the line through \((a, 0)\) is horizontal. The general equation of the line can be given as:
\[ y = mx \quad \text{(where \(m\) is the slope)} \]
For the two points, we can express the equation as:
\[ x + y = a \]

Thus, the equation of the line joining the origin to the points of intersection is:
\[ x + y = a \]