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. `(-3,6)`

Answer

To solve the problem, we need to find the center of circle S that passes through the intersection points of the two given circles and lies on the specified line. We will follow these steps:

**Step 1: Write the equations of the given circles.**

The equations of the circles are:
1. \( x^2 + y^2 - 6x = 0 \)
2. \( x^2 + y^2 - 4y = 0 \)

**Step 2: Rewrite the equations in standard form.**

For the first circle:
\[ x^2 - 6x + y^2 = 0 \implies (x - 3)^2 + y^2 = 9 \]
This represents a circle with center (3, 0) and radius 3.

For the second circle:
\[ x^2 + y^2 - 4y = 0 \implies x^2 + (y - 2)^2 = 4 \]
This represents a circle with center (0, 2) and radius 2.

**Step 3: Find the intersection points of the two circles.**

To find the intersection points, we can set the equations equal to each other. From the first circle, we have:
\[ y^2 = 6x - x^2 \]
Substituting this into the second circle's equation:
\[ x^2 + (6x - x^2 - 4y) = 0 \]
This simplifies to:
\[ x^2 + (6x - x^2 - 4(2 - \sqrt{4 - x^2})) = 0 \]
Solving this will give us the intersection points.

**Step 4: Use the family of circles method.**

The general form of the family of circles passing through the intersection of the two given circles is:
\[ S_1 + \lambda S_2 = 0 \]
Where \(S_1\) and \(S_2\) are the equations of the two circles. Thus,
\[ (x^2 + y^2 - 6x) + \lambda (x^2 + y^2 - 4y) = 0 \]
This simplifies to:
\[ (1 + \lambda)x^2 + (1 + \lambda)y^2 - (6 + 4\lambda)x - 4\lambda y = 0 \]

**Step 5: Identify the center of circle S.**

From the above equation, we can identify the center of circle S as:
\[ \left(\frac{6 + 4\lambda}{2(1 + \lambda)}, \frac{4\lambda}{2(1 + \lambda)}\right) = \left(\frac{3 + 2\lambda}{1 + \lambda}, \frac{2\lambda}{1 + \lambda}\right) \]

**Step 6: Substitute the center into the line equation.**

The center lies on the line \(2x - 3y + 12 = 0\). Substituting the center coordinates:
\[ 2\left(\frac{3 + 2\lambda}{1 + \lambda}\right) - 3\left(\frac{2\lambda}{1 + \lambda}\right) + 12 = 0 \]
Solving this equation will give us the value of \(\lambda\).

**Step 7: Find the equation of circle S.**

Once we have \(\lambda\), we can substitute it back into the equation of circle S to find its specific equation.

**Step 8: Determine the points through which circle S passes.**

Finally, we can test the given options to see which point lies on the circle S by substituting the coordinates of each point into the equation of circle S.

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