The maximum distance of the point (4,4) from the circle `x^2+y^2-2x-15=0` is

Options

A. 10
B. 9
C. 5
D. none of these

Answer

To find the maximum distance of the point (4, 4) from the circle given by the equation \(x^2 + y^2 - 2x - 15 = 0\), we will follow these steps:

### Step 1: Rewrite the Circle's Equation
First, we need to rewrite the equation of the circle in standard form. The given equation is:
\[ x^2 + y^2 - 2x - 15 = 0 \]
We can rearrange it as:
\[ x^2 - 2x + y^2 = 15 \]
Now, we complete the square for the \(x\) terms:
\[ (x^2 - 2x + 1) + y^2 = 15 + 1 \]
This simplifies to:
\[ (x - 1)^2 + y^2 = 16 \]
This means the circle has a center at \(C(1, 0)\) and a radius \(R = \sqrt{16} = 4\).

### Step 2: Determine the Location of the Point Relative to the Circle
Next, we will check whether the point (4, 4) lies inside, on, or outside the circle. We can do this by calculating the value of \(S_1\) using the formula:
\[ S_1 = (x - 1)^2 + (y - 0)^2 - R^2 \]
Substituting \(x = 4\) and \(y = 4\):
\[ S_1 = (4 - 1)^2 + (4 - 0)^2 - 16 \]
Calculating this gives:
\[ S_1 = 3^2 + 4^2 - 16 = 9 + 16 - 16 = 9 \]
Since \(S_1 > 0\), the point (4, 4) lies outside the circle.

### Step 3: Calculate the Distance from the Point to the Center of the Circle
Now we will find the distance from the point (4, 4) to the center of the circle (1, 0). The distance \(d\) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates:
\[ d = \sqrt{(4 - 1)^2 + (4 - 0)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

### Step 4: Calculate the Maximum Distance
The maximum distance from the point (4, 4) to the circle is the distance from the point to the center of the circle plus the radius of the circle:
\[ \text{Maximum Distance} = d + R = 5 + 4 = 9 \]

### Final Answer
Thus, the maximum distance of the point (4, 4) from the circle is:
\[ \boxed{9} \]