The maximum distance of the point (4,4) from the circle `x^2+y2-2x-15=0` is

Options

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

Answer

To find the maximum distance from the point (4, 4) to 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:
\[ (x - 1)^2 + y^2 = 16 \]
Thus, the center of the circle \(C\) is at \((1, 0)\) and the radius \(r\) is \(4\) (since \(\sqrt{16} = 4\)).

### Step 2: Calculate the Distance from the Point to the Center of the Circle
Next, we calculate the distance \(CP\) from the point \(P(4, 4)\) to the center \(C(1, 0)\) of the circle. The distance formula 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 in our points:
\[ CP = \sqrt{(4 - 1)^2 + (4 - 0)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

### Step 3: Find the Maximum Distance
Since the point \(P(4, 4)\) is outside the circle, the maximum distance from the point to the circle is given by:
\[ \text{Maximum Distance} = CP + r \]
Substituting the values we found:
\[ \text{Maximum Distance} = 5 + 4 = 9 \]

### Final Answer
The maximum distance of the point (4, 4) from the circle is \(9\).

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