The equation of the circle having centre (0, 0) and passing through the point of intersection of the lines `4x + 3y = 2 and x + 2y = 3` is

Options

A. `x^(2) + y^(2) = sqrt(5)`
B. `x^(2) + y^(2) = 5`
C. `x^(2) + y^(2) = 4`
D. `x^(2) + y^(2) = 2`

Answer

To find the equation of the circle with center at (0, 0) that passes through the point of intersection of the lines \(4x + 3y = 2\) and \(x + 2y = 3\), we can follow these steps:

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

We can solve these equations simultaneously. Let's express \(x\) from Equation 2:
\[ x = 3 - 2y \]
Now, substitute this value of \(x\) into Equation 1:
\[ 4(3 - 2y) + 3y = 2 \]
Expanding this gives:
\[ 12 - 8y + 3y = 2 \]
Combining like terms:
\[ 12 - 5y = 2 \]
Now, isolate \(y\):
\[ -5y = 2 - 12 \]
\[ -5y = -10 \]
\[ y = 2 \]

Now substitute \(y = 2\) back into Equation 2 to find \(x\):
\[ x + 2(2) = 3 \]
\[ x + 4 = 3 \]
\[ x = 3 - 4 = -1 \]

So, the point of intersection \(Q\) is \((-1, 2)\).

### Step 2: Find the radius of the circle
The radius \(r\) of the circle is the distance from the center (0, 0) to the point \(Q(-1, 2)\). We can use the distance formula:
\[ r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates:
\[ r = \sqrt{(-1 - 0)^2 + (2 - 0)^2} \]
\[ r = \sqrt{(-1)^2 + (2)^2} \]
\[ r = \sqrt{1 + 4} = \sqrt{5} \]

### Step 3: Write the equation of the circle
The general equation of a circle with center at the origin (0, 0) and radius \(r\) is given by:
\[ x^2 + y^2 = r^2 \]
Substituting \(r = \sqrt{5}\):
\[ x^2 + y^2 = (\sqrt{5})^2 \]
\[ x^2 + y^2 = 5 \]

### Final Answer
The equation of the circle is:
\[ \boxed{x^2 + y^2 = 5} \]