if `y = mx` is a chord of a circle of radius `a` and the diameter of the circle lies along `x`-axis and one end of this chord in origin .The equation of the circle described on this chord as diameter is

Options

A. Option1. `(1+m^(2))(x^(2)+y^(2))-2a(x+my)=0`
B. Option2. `(1-m^(2))(x^(2)+y^(2))-2a(x+my)=0`
C. Option3. `(1+m^(2))(x^(2)+y^(2))+2a(x+my)=0`
D. Option4. none of these

Answer

To find the equation of the circle described on the chord \( y = mx \) as the diameter, we will follow these steps:

### Step 1: Understand the Circle's Properties
The circle has a radius \( a \) and its diameter lies along the x-axis. The center of the circle will be at the point \( (a, 0) \) since the radius is \( a \).

### Step 2: Write the Equation of the Circle
The standard equation of a circle with center \( (h, k) \) and radius \( r \) is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
For our circle, substituting \( h = a \), \( k = 0 \), and \( r = a \):
\[ (x - a)^2 + y^2 = a^2 \]

### Step 3: Simplify the Equation
Expanding the equation:
\[ (x - a)^2 + y^2 = a^2 \]
\[ x^2 - 2ax + a^2 + y^2 = a^2 \]
Subtracting \( a^2 \) from both sides:
\[ x^2 - 2ax + y^2 = 0 \]

### Step 4: Find the Intersection Points with the Chord
The chord is given by \( y = mx \). Substituting \( y = mx \) into the circle's equation:
\[ x^2 - 2ax + (mx)^2 = 0 \]
\[ x^2 - 2ax + m^2x^2 = 0 \]
Factoring out \( x \):
\[ x(1 + m^2)x - 2a = 0 \]
This gives us:
\[ x = 0 \quad \text{or} \quad x = \frac{2a}{1 + m^2} \]

### Step 5: Find the Corresponding y-coordinates
For \( x = \frac{2a}{1 + m^2} \):
\[ y = m \left(\frac{2a}{1 + m^2}\right) = \frac{2am}{1 + m^2} \]

### Step 6: Find the Midpoint of the Chord
The endpoints of the chord are \( (0, 0) \) and \( \left(\frac{2a}{1 + m^2}, \frac{2am}{1 + m^2}\right) \). The midpoint \( M \) of the chord can be calculated as:
\[ M = \left(\frac{0 + \frac{2a}{1 + m^2}}{2}, \frac{0 + \frac{2am}{1 + m^2}}{2}\right) = \left(\frac{a}{1 + m^2}, \frac{am}{1 + m^2}\right) \]

### Step 7: Find the Radius of the Circle on the Chord
The radius \( r \) of the circle described on the chord as diameter is half the distance between the endpoints:
\[ r = \frac{1}{2} \sqrt{\left(\frac{2a}{1 + m^2}\right)^2 + \left(\frac{2am}{1 + m^2}\right)^2} \]
Calculating this gives:
\[ r = \frac{a}{\sqrt{1 + m^2}} \]

### Step 8: Write the Equation of the Circle with the Chord as Diameter
Now, the center of the circle is at \( M \) and the radius is \( r \):
\[ \left(x - \frac{a}{1 + m^2}\right)^2 + \left(y - \frac{am}{1 + m^2}\right)^2 = \left(\frac{a}{\sqrt{1 + m^2}}\right)^2 \]
This simplifies to:
\[ \left(x - \frac{a}{1 + m^2}\right)^2 + \left(y - \frac{am}{1 + m^2}\right)^2 = \frac{a^2}{1 + m^2} \]

### Final Equation
The final equation of the circle described on the chord as diameter is:
\[ \left(x - \frac{a}{1 + m^2}\right)^2 + \left(y - \frac{am}{1 + m^2}\right)^2 = \frac{a^2}{1 + m^2} \]