Find the equation of the circle whose centre is the point of intersection of the lines `2x-3y+4=0and3x+4y-5=0` and passes through the origin.

Answer

To find the equation of the circle whose center is the point of intersection of the lines \(2x - 3y + 4 = 0\) and \(3x + 4y - 5 = 0\) and which passes through the origin, we will follow these steps:

### Step 1: Find the point of intersection of the two lines.
We have the equations:
1. \(2x - 3y + 4 = 0\) (Equation 1)
2. \(3x + 4y - 5 = 0\) (Equation 2)

To find the intersection, we can eliminate one variable. Let's eliminate \(x\).

Multiply Equation 1 by 3:
\[ 6x - 9y + 12 = 0 \quad (3 \times \text{Equation 1}) \]

Multiply Equation 2 by 2:
\[ 6x + 8y - 10 = 0 \quad (2 \times \text{Equation 2}) \]

Now we have:
1. \(6x - 9y + 12 = 0\)
2. \(6x + 8y - 10 = 0\)

### Step 2: Subtract the two equations.
Subtract the second equation from the first:
\[ (6x - 9y + 12) - (6x + 8y - 10) = 0 \]
This simplifies to:
\[ -17y + 22 = 0 \]
So,
\[ 17y = 22 \implies y = \frac{22}{17} \]

### Step 3: Substitute \(y\) back to find \(x\).
Now substitute \(y = \frac{22}{17}\) back into Equation 1:
\[ 2x - 3\left(\frac{22}{17}\right) + 4 = 0 \]
This simplifies to:
\[ 2x - \frac{66}{17} + 4 = 0 \]
Convert 4 to a fraction:
\[ 2x - \frac{66}{17} + \frac{68}{17} = 0 \]
Combine the fractions:
\[ 2x + \frac{2}{17} = 0 \implies 2x = -\frac{2}{17} \implies x = -\frac{1}{17} \]

### Step 4: Determine the center of the circle.
The center of the circle is at the point:
\[ \left(-\frac{1}{17}, \frac{22}{17}\right) \]

### Step 5: Use the center to write the equation of the circle.
The general equation of a circle with center \((h, k)\) is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting \(h = -\frac{1}{17}\) and \(k = \frac{22}{17}\):
\[ \left(x + \frac{1}{17}\right)^2 + \left(y - \frac{22}{17}\right)^2 = r^2 \]

### Step 6: Find the radius \(r\) using the fact that the circle passes through the origin \((0, 0)\).
Substituting \((0, 0)\) into the circle equation:
\[ \left(0 + \frac{1}{17}\right)^2 + \left(0 - \frac{22}{17}\right)^2 = r^2 \]
Calculating:
\[ \left(\frac{1}{17}\right)^2 + \left(-\frac{22}{17}\right)^2 = r^2 \]
\[ \frac{1}{289} + \frac{484}{289} = r^2 \]
\[ \frac{485}{289} = r^2 \]

### Step 7: Write the final equation of the circle.
Thus, the equation of the circle is:
\[ \left(x + \frac{1}{17}\right)^2 + \left(y - \frac{22}{17}\right)^2 = \frac{485}{289} \]