A right angled isosceles triangle is inscribed in the circle `x^(2)+y^(2)-4x-2y-4=0` then length of the side of the triangle is

Options

A. `sqrt(2)`
B. `2sqrt(2)`
C. `3sqrt(2)`
D. `5sqrt(2)`

Answer

To solve the problem of finding the length of the side of the right-angled isosceles triangle inscribed in the given circle, we will follow these steps:

### Step 1: Rewrite the equation of the circle in standard form.
The given equation of the circle is:
\[ x^2 + y^2 - 4x - 2y - 4 = 0 \]

To rewrite this in standard form, we will complete the square for both \(x\) and \(y\).

1. For \(x\):
\[ x^2 - 4x \]
Completing the square:
\[ (x - 2)^2 - 4 \]

2. For \(y\):
\[ y^2 - 2y \]
Completing the square:
\[ (y - 1)^2 - 1 \]

Now substituting back into the equation:
\[ (x - 2)^2 - 4 + (y - 1)^2 - 1 - 4 = 0 \]
\[ (x - 2)^2 + (y - 1)^2 - 9 = 0 \]
\[ (x - 2)^2 + (y - 1)^2 = 9 \]

This shows that the circle is centered at \((2, 1)\) with a radius \(r = 3\).

### Step 2: Relate the triangle to the circle.
In a right-angled isosceles triangle, the hypotenuse is equal to the diameter of the circumcircle. The diameter \(D\) of the circle is twice the radius:
\[ D = 2r = 2 \times 3 = 6 \]

### Step 3: Use the properties of the triangle.
For a right-angled isosceles triangle, if the length of each equal side is \(a\), then the length of the hypotenuse \(h\) can be expressed as:
\[ h = a\sqrt{2} \]

Since we have established that the hypotenuse is equal to the diameter of the circle:
\[ a\sqrt{2} = 6 \]

### Step 4: Solve for \(a\).
To find \(a\), we rearrange the equation:
\[ a = \frac{6}{\sqrt{2}} = 6 \cdot \frac{\sqrt{2}}{2} = 3\sqrt{2} \]

### Conclusion:
The length of the side of the triangle is:
\[ \boxed{3\sqrt{2}} \]