Equation of the circle passing through the three points
(0, 0) , (0,b) and (a,b) is

Options

A. `x^(2) + y^(2) + ax + by = 0 `
B. `x^(2) + y^(2) - 6x + 12y - 15 = 0 `
C. `x^(2) + y^(2) - ax - by = 0 `
D. `x^(2) + y^(2) - 6x + 12y + 45 = 0 `

Answer

To find the equation of the circle passing through the points (0, 0), (0, b), and (a, b), we can follow these steps:

### Step 1: General Equation of a Circle
The general equation of a circle in the Cartesian plane is given by:
\[ x^2 + y^2 + 2Gx + 2Fy + C = 0 \]
where \(G\), \(F\), and \(C\) are constants that we need to determine.

### Step 2: Substitute the First Point (0, 0)
Substituting the point (0, 0) into the equation:
\[ 0^2 + 0^2 + 2G(0) + 2F(0) + C = 0 \]
This simplifies to:
\[ C = 0 \]

### Step 3: Substitute the Second Point (0, b)
Now, substituting the point (0, b):
\[ 0^2 + b^2 + 2G(0) + 2Fb + C = 0 \]
Since \(C = 0\), we have:
\[ b^2 + 2Fb = 0 \]
Factoring out \(b\) (assuming \(b \neq 0\)):
\[ b(2F + b) = 0 \]
Thus, we get:
\[ 2F + b = 0 \implies F = -\frac{b}{2} \]

### Step 4: Substitute the Third Point (a, b)
Next, we substitute the point (a, b):
\[ a^2 + b^2 + 2Ga + 2Fb + C = 0 \]
Again, since \(C = 0\) and \(F = -\frac{b}{2}\), we have:
\[ a^2 + b^2 + 2Ga + 2\left(-\frac{b}{2}\right)b = 0 \]
This simplifies to:
\[ a^2 + b^2 + 2Ga - b^2 = 0 \]
Thus:
\[ a^2 + 2Ga = 0 \]
Factoring out \(a\) (assuming \(a \neq 0\)):
\[ a(2G + a) = 0 \]
Therefore:
\[ 2G + a = 0 \implies G = -\frac{a}{2} \]

### Step 5: Substitute G, F, and C into the General Equation
Now we substitute \(G\), \(F\), and \(C\) back into the general equation of the circle:
\[ x^2 + y^2 + 2\left(-\frac{a}{2}\right)x + 2\left(-\frac{b}{2}\right)y + 0 = 0 \]
This simplifies to:
\[ x^2 + y^2 - ax - by = 0 \]

### Final Equation
Thus, the equation of the circle passing through the points (0, 0), (0, b), and (a, b) is:
\[ x^2 + y^2 - ax - by = 0 \]

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