Find the equation of the circle
(i) whose centre is (a,b) which passes through the origin
(ii) whose centre is the point (2,3) and which passes through the intersection of the lines 3x-2y-1=0 and 4x+y-27 =0

Answer

To solve the given problem, we will find the equations of two circles based on the provided conditions.

### (i) Finding the equation of the circle with center (a, b) that passes through the origin:

1. **General Equation of a Circle**:
The general equation of a circle with center (h, k) and radius r is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Here, the center is (a, b), so we substitute h = a and k = b:
\[ (x - a)^2 + (y - b)^2 = r^2 \]

2. **Condition of Passing Through the Origin**:
Since the circle passes through the origin (0, 0), we substitute x = 0 and y = 0 into the equation:
\[ (0 - a)^2 + (0 - b)^2 = r^2 \]
This simplifies to:
\[ a^2 + b^2 = r^2 \]

3. **Substituting r² in the Circle Equation**:
We can replace \( r^2 \) in the circle's equation with \( a^2 + b^2 \):
\[ (x - a)^2 + (y - b)^2 = a^2 + b^2 \]

4. **Final Equation of the First Circle**:
The equation of the first circle is:
\[ (x - a)^2 + (y - b)^2 = a^2 + b^2 \]

### (ii) Finding the equation of the circle with center (2, 3) that passes through the intersection of the lines:

1. **Equation of the Circle**:
The center of the circle is (2, 3). Thus, the general equation becomes:
\[ (x - 2)^2 + (y - 3)^2 = r^2 \]

2. **Finding the Intersection of the Lines**:
We need to find the intersection of the lines given by:
\[ 3x - 2y - 1 = 0 \quad \text{(Equation 1)} \]
\[ 4x + y - 27 = 0 \quad \text{(Equation 2)} \]

From Equation 2, we can express y in terms of x:
\[ y = 27 - 4x \]

Now, substitute this value of y into Equation 1:
\[ 3x - 2(27 - 4x) - 1 = 0 \]
Simplifying this:
\[ 3x - 54 + 8x - 1 = 0 \]
\[ 11x - 55 = 0 \]
\[ x = 5 \]

3. **Finding y-coordinate**:
Substitute \( x = 5 \) back into Equation 2 to find y:
\[ y = 27 - 4(5) = 27 - 20 = 7 \]
Thus, the point of intersection is (5, 7).

4. **Finding r²**:
Now, we substitute the point (5, 7) into the circle's equation to find r²:
\[ (5 - 2)^2 + (7 - 3)^2 = r^2 \]
\[ 3^2 + 4^2 = r^2 \]
\[ 9 + 16 = r^2 \]
\[ r^2 = 25 \]

5. **Final Equation of the Second Circle**:
The equation of the second circle is:
\[ (x - 2)^2 + (y - 3)^2 = 25 \]

### Summary of the Results:
- The equation of the first circle is:
\[ (x - a)^2 + (y - b)^2 = a^2 + b^2 \]
- The equation of the second circle is:
\[ (x - 2)^2 + (y - 3)^2 = 25 \]