Find the equation of the circle which passes through the origin and belonges to the co-axial of circles whose limiting points are (1,2) and (4,3).

Answer

To find the equation of the circle that passes through the origin and belongs to the coaxial family of circles with limiting points at (1, 2) and (4, 3), we can follow these steps:

### Step 1: Write the equations of the circles with given limiting points.
The limiting points of the circles are (1, 2) and (4, 3). The equations of the circles with these limiting points (having radius 0) can be written as:

1. For point (1, 2):
\[ (x - 1)^2 + (y - 2)^2 = 0 \]
Expanding this gives:
\[ x^2 - 2x + 1 + y^2 - 4y + 4 = 0 \implies x^2 + y^2 - 2x - 4y + 5 = 0 \]

2. For point (4, 3):
\[ (x - 4)^2 + (y - 3)^2 = 0 \]
Expanding this gives:
\[ x^2 - 8x + 16 + y^2 - 6y + 9 = 0 \implies x^2 + y^2 - 8x - 6y + 25 = 0 \]

### Step 2: Formulate the family of coaxial circles.
The family of coaxial circles can be expressed as:
\[ C_1 + \lambda C_2 = 0 \]
Where \(C_1\) and \(C_2\) are the equations we derived above. Therefore, we have:
\[ (x^2 + y^2 - 2x - 4y + 5) + \lambda (x^2 + y^2 - 8x - 6y + 25) = 0 \]

### Step 3: Combine the equations.
Combining the equations gives:
\[ (1 + \lambda)x^2 + (1 + \lambda)y^2 + (-2 - 8\lambda)x + (-4 - 6\lambda)y + (5 + 25\lambda) = 0 \]

### Step 4: Condition for passing through the origin.
For the circle to pass through the origin (0, 0), we substitute \(x = 0\) and \(y = 0\) into the equation:
\[ 5 + 25\lambda = 0 \]
Solving for \(\lambda\):
\[ 25\lambda = -5 \implies \lambda = -\frac{1}{5} \]

### Step 5: Substitute \(\lambda\) back into the equation.
Now substituting \(\lambda = -\frac{1}{5}\) into the combined equation:
\[ (1 - \frac{1}{5})x^2 + (1 - \frac{1}{5})y^2 + (-2 + \frac{8}{5})x + (-4 + \frac{6}{5})y + (5 - 5) = 0 \]
This simplifies to:
\[ \frac{4}{5}x^2 + \frac{4}{5}y^2 - \frac{2}{5}x - \frac{14}{5}y = 0 \]

### Step 6: Clear the fractions by multiplying through by 5:
Multiplying through by 5 gives:
\[ 4x^2 + 4y^2 - 2x - 14y = 0 \]
Dividing by 2:
\[ 2x^2 + 2y^2 - x - 7y = 0 \]

### Final Equation:
Thus, the equation of the circle is:
\[ 2x^2 + 2y^2 - x - 7y = 0 \]