If the line 3x-2y + 6=0 meets X-axis and Y-axis respectively at A and B, then find the equation of the circle with radius AB and centre at A.

Options

A. `x^(2)+y^(2)+4x+9=0`
B. `x^(2)+y^(2)+4x-9=0`
C. `x^(2)+y^(2)-4+4=0`
D. `x^(2)+y^(2)+4x-4=0`

Answer

To solve the problem, we will follow these steps:

### Step 1: Find the coordinates of points A and B
The line given is \(3x - 2y + 6 = 0\). To find where this line intersects the x-axis (point A), we set \(y = 0\):

\[ 3x + 6 = 0 \implies 3x = -6 \implies x = -2 \]

So, the coordinates of point A are \((-2, 0)\).

Next, to find where this line intersects the y-axis (point B), we set \(x = 0\):

\[ -2y + 6 = 0 \implies -2y = -6 \implies y = 3 \]

Thus, the coordinates of point B are \((0, 3)\).

### Step 2: Calculate the distance AB
The distance \(AB\) can be calculated using the distance formula:

\[ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Substituting the coordinates of A and B:

\[ AB = \sqrt{(0 - (-2))^2 + (3 - 0)^2} = \sqrt{(0 + 2)^2 + 3^2} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \]

### Step 3: Write the equation of the circle
The center of the circle is at point A \((-2, 0)\) and the radius \(R\) is \(AB = \sqrt{13}\).

The standard equation of a circle with center \((h, k)\) and radius \(R\) is given by:

\[ (x - h)^2 + (y - k)^2 = R^2 \]

Substituting \(h = -2\), \(k = 0\), and \(R = \sqrt{13}\):

\[ (x - (-2))^2 + (y - 0)^2 = (\sqrt{13})^2 \]

This simplifies to:

\[ (x + 2)^2 + y^2 = 13 \]

### Step 4: Rearranging the equation
To express this in a standard form, we can expand it:

\[ (x + 2)^2 + y^2 = 13 \]

Expanding \((x + 2)^2\):

\[ x^2 + 4x + 4 + y^2 = 13 \]

Now, rearranging gives us:

\[ x^2 + 4x + y^2 + 4 - 13 = 0 \implies x^2 + 4x + y^2 - 9 = 0 \]

### Final Answer
The equation of the circle is:

\[ x^2 + 4x + y^2 - 9 = 0 \]

---