The least length of chord passing through (2,1) of the circle `x^(2)+y^(2)-2x-4y-13=0` is

Options

A. 2
B. 6
C. 8
D. 4

Answer

To find the least length of the chord passing through the point (2, 1) of the circle given by the equation \(x^2 + y^2 - 2x - 4y - 13 = 0\), we will follow these steps:

### Step 1: Rewrite the equation of the circle in standard form
The given equation of the circle is:
\[ x^2 + y^2 - 2x - 4y - 13 = 0 \]
We can rewrite this by completing the square for both \(x\) and \(y\).

For \(x\):
\[ x^2 - 2x \quad \text{can be rewritten as} \quad (x-1)^2 - 1 \]

For \(y\):
\[ y^2 - 4y \quad \text{can be rewritten as} \quad (y-2)^2 - 4 \]

Substituting these back into the equation:
\[ (x-1)^2 - 1 + (y-2)^2 - 4 - 13 = 0 \]
This simplifies to:
\[ (x-1)^2 + (y-2)^2 - 18 = 0 \]
Thus, the equation of the circle in standard form is:
\[ (x-1)^2 + (y-2)^2 = 18 \]

### Step 2: Identify the center and radius of the circle
From the standard form \((x-h)^2 + (y-k)^2 = r^2\), we can identify:
- Center \((h, k) = (1, 2)\)
- Radius \(r = \sqrt{18} = 3\sqrt{2}\)

### Step 3: Calculate the distance from the center to the point (2, 1)
Next, we need to find the distance from the center of the circle \((1, 2)\) to the point \((2, 1)\):
\[ d = \sqrt{(2-1)^2 + (1-2)^2} = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \]

### Step 4: Use the relationship between radius, distance, and chord length
The length of the chord \(AB\) passing through the point \((2, 1)\) can be calculated using the formula:
\[ AB = 2 \sqrt{r^2 - d^2} \]
Where:
- \(r = 3\sqrt{2}\)
- \(d = \sqrt{2}\)

Calculating \(r^2\) and \(d^2\):
\[ r^2 = (3\sqrt{2})^2 = 18 \]
\[ d^2 = (\sqrt{2})^2 = 2 \]

Now substituting into the chord length formula:
\[ AB = 2 \sqrt{18 - 2} = 2 \sqrt{16} = 2 \times 4 = 8 \]

### Conclusion
The least length of the chord passing through the point \((2, 1)\) is \(8\).