Center of a circle S passing through the intersection points of circles `x^2+y^2-6x=0 &x^2+y^2-4y=0` lies on the line `2x-3y+12=0` then circle S passes through

Options

A. `(-3,1)`
B. `(-4,-2)`
C. `(1,2)`
D. `(0,0)`

Answer

To solve the problem, we need to find the center of the circle \( S \) that passes through the intersection points of the two given circles and lies on the specified line. Here are the steps to arrive at the solution:

### Step 1: Write the equations of the circles
The equations of the two circles are given as:
1. \( x^2 + y^2 - 6x = 0 \)
2. \( x^2 + y^2 - 4y = 0 \)

### Step 2: Rearrange the equations
We can rearrange these equations to standard form:
1. \( (x - 3)^2 + y^2 = 9 \) (Circle with center (3, 0) and radius 3)
2. \( x^2 + (y - 2)^2 = 4 \) (Circle with center (0, 2) and radius 2)

### Step 3: Find the intersection points of the circles
To find the intersection points, we can set the two equations equal to each other:
\[ x^2 + y^2 - 6x = x^2 + y^2 - 4y \]
This simplifies to:
\[ -6x + 4y = 0 \implies 4y = 6x \implies y = \frac{3}{2}x \]

### Step 4: Substitute \( y \) in one of the circle equations
Now, substitute \( y = \frac{3}{2}x \) into one of the circle equations. Let's use the first circle:
\[ x^2 + \left(\frac{3}{2}x\right)^2 - 6x = 0 \]
This becomes:
\[ x^2 + \frac{9}{4}x^2 - 6x = 0 \]
Combining like terms:
\[ \frac{13}{4}x^2 - 6x = 0 \]
Factoring out \( x \):
\[ x\left(\frac{13}{4}x - 6\right) = 0 \]
Thus, \( x = 0 \) or \( \frac{13}{4}x - 6 = 0 \) which gives \( x = \frac{24}{13} \).

### Step 5: Find corresponding \( y \) values
For \( x = 0 \):
\[ y = \frac{3}{2}(0) = 0 \implies (0, 0) \]
For \( x = \frac{24}{13} \):
\[ y = \frac{3}{2} \cdot \frac{24}{13} = \frac{36}{13} \implies \left(\frac{24}{13}, \frac{36}{13}\right) \]

### Step 6: Find the center of circle \( S \)
The center of circle \( S \) can be expressed as:
\[ \left( \frac{3}{1 + \lambda}, \frac{2\lambda}{1 + \lambda} \right) \]
where \( \lambda \) is a parameter.

### Step 7: Substitute the center into the line equation
The center lies on the line \( 2x - 3y + 12 = 0 \):
\[ 2\left(\frac{3}{1 + \lambda}\right) - 3\left(\frac{2\lambda}{1 + \lambda}\right) + 12 = 0 \]
Multiplying through by \( 1 + \lambda \):
\[ 6 - 6\lambda + 12(1 + \lambda) = 0 \]
This simplifies to:
\[ 6 - 6\lambda + 12 + 12\lambda = 0 \implies 6 + 6\lambda = 0 \implies \lambda = -1 \]

### Step 8: Substitute \( \lambda \) back to find the circle equation
Substituting \( \lambda = -1 \):
\[ x^2 + y^2 - 6x + 4y = 0 \]

### Step 9: Check which points lie on the circle
Now we need to check which of the given points lie on the circle:
1. \( (-3, 1) \)
2. \( (-4, -2) \)
3. \( (1, 2) \)
4. \( (0, 0) \)

Substituting each point into the equation \( x^2 + y^2 - 6x + 4y = 0 \) to see which satisfies the equation.

### Final Answer
After checking, we find that the point \( (0, 0) \) satisfies the equation.