The minimum length of the chord of the circle `x^2+y^2+2x+2y-7=0` which is passing (1,0) is :

Options

A. 2
B. 4
C. `2sqrt(2)`
D. `sqrt(5)`

Answer

To find the minimum length of the chord of the circle \(x^2 + y^2 + 2x + 2y - 7 = 0\) that passes through the point (1, 0), we can follow these steps:

### Step 1: Rewrite the equation of the circle in standard form
The given equation of the circle is:
\[ x^2 + y^2 + 2x + 2y - 7 = 0 \]
We can complete the square for both \(x\) and \(y\).

For \(x\):
\[ x^2 + 2x = (x + 1)^2 - 1 \]

For \(y\):
\[ y^2 + 2y = (y + 1)^2 - 1 \]

Substituting these into the equation:
\[ (x + 1)^2 - 1 + (y + 1)^2 - 1 - 7 = 0 \]
\[ (x + 1)^2 + (y + 1)^2 - 9 = 0 \]
\[ (x + 1)^2 + (y + 1)^2 = 9 \]

This shows that the circle is centered at \((-1, -1)\) with a radius of \(3\).

### Step 2: Find the distance from the center of the circle to the point (1, 0)
The center of the circle is \((-1, -1)\) and the point through which the chord passes is \((1, 0)\). We can calculate the distance \(OA\) from the center \(O(-1, -1)\) to the point \(A(1, 0)\) using the distance formula:
\[ OA = \sqrt{(1 - (-1))^2 + (0 - (-1))^2} = \sqrt{(1 + 1)^2 + (0 + 1)^2} = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5} \]

### Step 3: Use the Pythagorean theorem to find the length of the chord
Let \(OB\) be the radius of the circle, which is \(3\), and \(AB\) be half the length of the chord. According to the Pythagorean theorem:
\[ OB^2 = OA^2 + AB^2 \]
Substituting the known values:
\[ 3^2 = (\sqrt{5})^2 + AB^2 \]
\[ 9 = 5 + AB^2 \]
\[ AB^2 = 9 - 5 = 4 \]
\[ AB = \sqrt{4} = 2 \]

### Step 4: Calculate the total length of the chord
Since \(AB\) is half the length of the chord, the total length of the chord \(CD\) is:
\[ CD = 2 \times AB = 2 \times 2 = 4 \]

### Final Answer
The minimum length of the chord of the circle that passes through the point (1, 0) is:
\[ \boxed{4} \]