If 'P' is any point on the circumference of the circle `x^(2) + y^(2) - 4x - 4y -8 =0` , then the perpendicular distance of the tangent at P from the centre of the circle is

Options

A. 8 units
B. 16 units
C. 4 units
D. 2 units

Answer

To find the perpendicular distance of the tangent at point P from the center of the circle given by the equation \( x^2 + y^2 - 4x - 4y - 8 = 0 \), we will follow these steps:

### Step 1: Rewrite the equation of the circle in standard form

The given equation of the circle is:
\[ x^2 + y^2 - 4x - 4y - 8 = 0 \]

We can rearrange this equation to find the center and radius. To do this, we will complete the square for both \(x\) and \(y\).

### Step 2: Completing the square

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

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

Now substituting back into the equation:
\[ (x - 2)^2 - 4 + (y - 2)^2 - 4 - 8 = 0 \]

This simplifies to:
\[ (x - 2)^2 + (y - 2)^2 - 16 = 0 \]

Thus, we have:
\[ (x - 2)^2 + (y - 2)^2 = 16 \]

### Step 3: Identify the center and radius

From the standard form of the circle \((x - h)^2 + (y - k)^2 = r^2\), we can identify:
- Center \(C(2, 2)\)
- Radius \(r = \sqrt{16} = 4\)

### Step 4: Find the perpendicular distance from the center to the tangent at point P

The tangent at any point \(P\) on the circle is perpendicular to the radius at that point. The perpendicular distance from the center \(C\) to the tangent line at point \(P\) is equal to the radius of the circle.

Thus, the perpendicular distance from the center of the circle to the tangent at point \(P\) is:
\[ \text{Distance} = r = 4 \]

### Final Answer

The perpendicular distance of the tangent at point P from the center of the circle is \(4\).

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