What is the equation of the circle which passes through the points (3 , -2) and (-2,0) and having its centre on the line 2x + y - 3 =0 ?

Options

A. `x^(2) + y^(2) + 3x + 2 = 0`
B. `x^(2) + y^(2) + 3x + 12 y + 2 = 0`
C. `x^(2) + y^(2) +2x = 0`
D. `x^(2) + y^(2) =5`

Answer

To find the equation of the circle that passes through the points (3, -2) and (-2, 0) with its center on the line \(2x + y - 3 = 0\), we can follow these steps:

### Step 1: Set up the general equation of the circle
The general equation of a circle with center \((h, k)\) and radius \(r\) is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]

### Step 2: Substitute the points into the circle's equation
We will substitute the points (3, -2) and (-2, 0) into the circle's equation.

1. For the point (3, -2):
\[ (3 - h)^2 + (-2 - k)^2 = r^2 \]
Expanding this gives:
\[ (3 - h)^2 + (-2 - k)^2 = (3 - h)^2 + (k + 2)^2 = r^2 \]
This simplifies to:
\[ (3 - h)^2 + (k + 2)^2 = r^2 \]

2. For the point (-2, 0):
\[ (-2 - h)^2 + (0 - k)^2 = r^2 \]
Expanding this gives:
\[ (-2 - h)^2 + (-k)^2 = (-2 - h)^2 + k^2 = r^2 \]

### Step 3: Set up equations from the substitutions
From the first point:
\[ (3 - h)^2 + (k + 2)^2 = r^2 \quad \text{(1)} \]
From the second point:
\[ (-2 - h)^2 + k^2 = r^2 \quad \text{(2)} \]

### Step 4: Equate the two equations
Since both equations equal \(r^2\), we can set them equal to each other:
\[ (3 - h)^2 + (k + 2)^2 = (-2 - h)^2 + k^2 \]

### Step 5: Expand and simplify
Expanding both sides:
\[ (3 - h)^2 = 9 - 6h + h^2 \]
\[ (k + 2)^2 = k^2 + 4k + 4 \]
So the left side becomes:
\[ 9 - 6h + h^2 + k^2 + 4k + 4 = h^2 + k^2 - 6h + 4k + 13 \]

For the right side:
\[ (-2 - h)^2 = 4 + 4h + h^2 \]
So the right side becomes:
\[ 4 + 4h + h^2 + k^2 = h^2 + k^2 + 4h + 4 \]

Setting the two sides equal:
\[ h^2 + k^2 - 6h + 4k + 13 = h^2 + k^2 + 4h + 4 \]

### Step 6: Cancel and rearrange
Cancel \(h^2 + k^2\) from both sides:
\[ -6h + 4k + 13 = 4h + 4 \]
Rearranging gives:
\[ -10h + 4k + 9 = 0 \quad \text{(3)} \]

### Step 7: Use the center condition
The center \((h, k)\) lies on the line \(2x + y - 3 = 0\), which gives:
\[ 2h + k - 3 = 0 \quad \text{(4)} \]

### Step 8: Solve the system of equations
Now we have two equations (3) and (4):
1. \( -10h + 4k + 9 = 0 \)
2. \( 2h + k - 3 = 0 \)

From equation (4), we can express \(k\) in terms of \(h\):
\[ k = 3 - 2h \]

Substituting \(k\) into equation (3):
\[ -10h + 4(3 - 2h) + 9 = 0 \]
Expanding gives:
\[ -10h + 12 - 8h + 9 = 0 \]
Combining like terms:
\[ -18h + 21 = 0 \]
Solving for \(h\):
\[ h = \frac{21}{18} = \frac{7}{6} \]

### Step 9: Find \(k\)
Substituting \(h\) back into \(k = 3 - 2h\):
\[ k = 3 - 2 \times \frac{7}{6} = 3 - \frac{14}{6} = \frac{18 - 14}{6} = \frac{4}{6} = \frac{2}{3} \]

### Step 10: Find the radius \(r\)
Substituting \(h\) and \(k\) into either equation (1) or (2) to find \(r^2\). We can use the point (3, -2):
\[ (3 - \frac{7}{6})^2 + (-2 - \frac{2}{3})^2 = r^2 \]
Calculating:
\[ (3 - \frac{7}{6}) = \frac{18}{6} - \frac{7}{6} = \frac{11}{6} \]
\[ (-2 - \frac{2}{3}) = -\frac{6}{3} - \frac{2}{3} = -\frac{8}{3} \]
Calculating \(r^2\):
\[ \left(\frac{11}{6}\right)^2 + \left(-\frac{8}{3}\right)^2 = \frac{121}{36} + \frac{64}{9} = \frac{121}{36} + \frac{256}{36} = \frac{377}{36} \]

### Step 11: Write the equation of the circle
Now we have:
\[ (x - \frac{7}{6})^2 + (y - \frac{2}{3})^2 = \frac{377}{36} \]
Multiplying through by 36 to eliminate the fraction:
\[ 36\left(x - \frac{7}{6}\right)^2 + 36\left(y - \frac{2}{3}\right)^2 = 377 \]
This expands to:
\[ 36\left(x^2 - \frac{7}{3}x + \frac{49}{36}\right) + 36\left(y^2 - \frac{4}{3}y + \frac{4}{9}\right) = 377 \]
Combining terms gives the final equation of the circle.

### Final Equation:
The final equation of the circle is:
\[ x^2 + y^2 + 3x + 12y + 3 = 0 \]