Equation of the circle passing through the origin and having its centre on the line `y = 3x` at a distance `sqrt10` from the origin is

Options

A. `x ^(2) + y ^(2) - 2x + 6y =0`
B. `x ^(2) + y ^(2) + 2x - 6y =0`
C. `x ^(2) + y ^(2) - 2x - 6y =0`
D. none of these

Answer

To find the equation of the circle that passes through the origin and has its center on the line \( y = 3x \) at a distance \( \sqrt{10} \) from the origin, we can follow these steps:

### Step 1: Determine the Center of the Circle
Let the center of the circle be \( (h, k) \). Since the center lies on the line \( y = 3x \), we can express \( k \) in terms of \( h \):
\[ k = 3h \]

### Step 2: Distance from the Origin
The distance from the origin to the center \( (h, k) \) is given as \( \sqrt{10} \). We can use the distance formula:
\[ \sqrt{h^2 + k^2} = \sqrt{10} \]
Squaring both sides, we have:
\[ h^2 + k^2 = 10 \]
Substituting \( k = 3h \) into the equation:
\[ h^2 + (3h)^2 = 10 \]
This simplifies to:
\[ h^2 + 9h^2 = 10 \]
\[ 10h^2 = 10 \]
\[ h^2 = 1 \]
Thus, we find:
\[ h = 1 \quad \text{or} \quad h = -1 \]

### Step 3: Find Corresponding \( k \) Values
Using \( h = 1 \):
\[ k = 3(1) = 3 \quad \Rightarrow \quad (h, k) = (1, 3) \]
Using \( h = -1 \):
\[ k = 3(-1) = -3 \quad \Rightarrow \quad (h, k) = (-1, -3) \]

### Step 4: Calculate the Radius
The radius \( r \) of the circle is the distance from the center to the origin, which we already know is \( \sqrt{10} \).

### Step 5: Write the Equation of the Circle
The general equation of a circle with center \( (h, k) \) and radius \( r \) is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
For the center \( (1, 3) \):
\[ (x - 1)^2 + (y - 3)^2 = 10 \]
Expanding this:
\[ (x^2 - 2x + 1) + (y^2 - 6y + 9) = 10 \]
Combining terms:
\[ x^2 + y^2 - 2x - 6y + 10 = 10 \]
Simplifying gives:
\[ x^2 + y^2 - 2x - 6y = 0 \]

For the center \( (-1, -3) \):
\[ (x + 1)^2 + (y + 3)^2 = 10 \]
Expanding this:
\[ (x^2 + 2x + 1) + (y^2 + 6y + 9) = 10 \]
Combining terms:
\[ x^2 + y^2 + 2x + 6y + 10 = 10 \]
Simplifying gives:
\[ x^2 + y^2 + 2x + 6y = 0 \]

### Final Equations
Thus, the equations of the circles are:
1. \( x^2 + y^2 - 2x - 6y = 0 \)
2. \( x^2 + y^2 + 2x + 6y = 0 \)

Since the question asks for the equation of the circle passing through the origin and having its center on the line \( y = 3x \) at a distance \( \sqrt{10} \), we can conclude that the correct answer is:
\[ \text{Equation of the circle: } x^2 + y^2 - 2x - 6y = 0 \]