Find the equation of the circle which passes through the points (1,-2) and (4,-3) and which has its centre on the straight line 3x+4y=5.

Answer

To find the equation of the circle that passes through the points (1, -2) and (4, -3) and has its center on the line given by the equation \(3x + 4y = 5\), we can follow these steps:

### Step 1: Set up the general equation of the circle
The general equation of a circle can be written as:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
where \((-g, -f)\) is the center of the circle.

### Step 2: Substitute the points into the circle equation
Since the circle passes through the points (1, -2) and (4, -3), we can substitute these points into the circle equation to create two equations.

**For the point (1, -2):**
\[ 1^2 + (-2)^2 + 2g(1) + 2f(-2) + c = 0 \]
This simplifies to:
\[ 1 + 4 + 2g - 4f + c = 0 \quad \Rightarrow \quad 2g - 4f + c = -5 \quad \text{(Equation 1)} \]

**For the point (4, -3):**
\[ 4^2 + (-3)^2 + 2g(4) + 2f(-3) + c = 0 \]
This simplifies to:
\[ 16 + 9 + 8g - 6f + c = 0 \quad \Rightarrow \quad 8g - 6f + c = -25 \quad \text{(Equation 2)} \]

### Step 3: Use the line equation for the center
The center of the circle \((-g, -f)\) lies on the line given by \(3x + 4y = 5\). Substituting \(-g\) for \(x\) and \(-f\) for \(y\) gives:
\[ 3(-g) + 4(-f) = 5 \quad \Rightarrow \quad -3g - 4f = 5 \quad \Rightarrow \quad 3g + 4f = -5 \quad \text{(Equation 3)} \]

### Step 4: Solve the system of equations
Now we have three equations:
1. \(2g - 4f + c = -5\) (Equation 1)
2. \(8g - 6f + c = -25\) (Equation 2)
3. \(3g + 4f = -5\) (Equation 3)

We can eliminate \(c\) by subtracting Equation 1 from Equation 2:
\[ (8g - 6f + c) - (2g - 4f + c) = -25 + 5 \]
This simplifies to:
\[ 6g - 2f = -20 \quad \Rightarrow \quad 3g - f = -10 \quad \text{(Equation 4)} \]

### Step 5: Solve Equations 3 and 4
Now we can solve Equations 3 and 4 together:
1. \(3g + 4f = -5\) (Equation 3)
2. \(3g - f = -10\) (Equation 4)

From Equation 4, we can express \(f\) in terms of \(g\):
\[ f = 3g + 10 \]
Substituting this into Equation 3:
\[ 3g + 4(3g + 10) = -5 \]
This simplifies to:
\[ 3g + 12g + 40 = -5 \quad \Rightarrow \quad 15g + 40 = -5 \quad \Rightarrow \quad 15g = -45 \quad \Rightarrow \quad g = -3 \]

### Step 6: Find \(f\) using \(g\)
Substituting \(g = -3\) back into Equation 4:
\[ 3(-3) - f = -10 \quad \Rightarrow \quad -9 - f = -10 \quad \Rightarrow \quad f = 1 \]

### Step 7: Find \(c\) using \(g\) and \(f\)
Now, substitute \(g\) and \(f\) into Equation 1 to find \(c\):
\[ 2(-3) - 4(1) + c = -5 \quad \Rightarrow \quad -6 - 4 + c = -5 \quad \Rightarrow \quad c = 5 \]

### Step 8: Write the final equation of the circle
Now we have \(g = -3\), \(f = 1\), and \(c = 5\). Substitute these values into the general circle equation:
\[ x^2 + y^2 + 2(-3)x + 2(1)y + 5 = 0 \]
This simplifies to:
\[ x^2 + y^2 - 6x + 2y + 5 = 0 \]

### Final Answer
The equation of the circle is:
\[ x^2 + y^2 - 6x + 2y + 5 = 0 \]