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

To find the intersection, we can solve these equations simultaneously.

From the first equation, we can express \(y\) in terms of \(x\):
\[ 3y = 2x + 4 \implies y = \frac{2x + 4}{3} \]

Now, substitute \(y\) in the second equation:
\[ 3x + 4\left(\frac{2x + 4}{3}\right) - 5 = 0 \]
Multiply through by 3 to eliminate the fraction:
\[ 9x + 4(2x + 4) - 15 = 0 \]
\[ 9x + 8x + 16 - 15 = 0 \]
\[ 17x + 1 = 0 \implies x = -\frac{1}{17} \]

Now substitute \(x\) back to find \(y\):
\[ y = \frac{2\left(-\frac{1}{17}\right) + 4}{3} = \frac{-\frac{2}{17} + \frac{68}{17}}{3} = \frac{\frac{66}{17}}{3} = \frac{22}{17} \]

Thus, the point of intersection (center of the circle) is:
\[ \left(-\frac{1}{17}, \frac{22}{17}\right) \]

### Step 2: Find the radius of the circle
The circle passes through the origin \((0, 0)\). The radius \(r\) can be calculated using the distance formula between the center and the origin:
\[ r = \sqrt{\left(-\frac{1}{17} - 0\right)^2 + \left(\frac{22}{17} - 0\right)^2} \]
\[ r = \sqrt{\left(-\frac{1}{17}\right)^2 + \left(\frac{22}{17}\right)^2} \]
\[ = \sqrt{\frac{1}{289} + \frac{484}{289}} = \sqrt{\frac{485}{289}} = \frac{\sqrt{485}}{17} \]

### Step 3: Write the equation of the circle
The standard form of the equation of a circle with center \((h, k)\) and radius \(r\) is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting \(h = -\frac{1}{17}\), \(k = \frac{22}{17}\), and \(r^2 = \frac{485}{289}\):
\[ \left(x + \frac{1}{17}\right)^2 + \left(y - \frac{22}{17}\right)^2 = \frac{485}{289} \]

To eliminate the fractions, we can multiply through by \(289\):
\[ 289\left(x + \frac{1}{17}\right)^2 + 289\left(y - \frac{22}{17}\right)^2 = 485 \]
This simplifies to:
\[ 17(x + 1/17)^2 + 17(y - 22/17)^2 = 485 \]
Thus, the final equation of the circle is:
\[ 17x + 1)^2 + 17(y - 22)^2 = 485 \]

### Final Answer
The required equation of the circle is:
\[ 17(x + \frac{1}{17})^2 + 17(y - \frac{22}{17})^2 = 485 \]