The power of (2,1) with respect to the circle `x^(2)+y^(2)+x=0` is

Options

A. 2
B. 3
C. `sqrt(7)`
D. 7

Answer

To find the power of the point (2, 1) with respect to the circle given by the equation \(x^2 + y^2 + x = 0\), we will follow these steps:

### Step 1: Rewrite the Circle Equation
The equation of the circle can be rewritten in standard form. We start with:
\[ x^2 + y^2 + x = 0 \]
We can rearrange this to:
\[ x^2 + x + y^2 = 0 \]
To complete the square for the \(x\) terms:
\[ x^2 + x = (x + \frac{1}{2})^2 - \frac{1}{4} \]
Thus, the equation becomes:
\[ (x + \frac{1}{2})^2 + y^2 = \frac{1}{4} \]
This shows that the circle has a center at \((-0.5, 0)\) and a radius of \(\frac{1}{2}\).

### Step 2: Substitute the Point into the Circle Equation
Let \(S\) be the equation of the circle:
\[ S = x^2 + y^2 + x \]
Now, substitute the coordinates of the point (2, 1) into this equation to find \(S_1\):
\[ S_1 = (2)^2 + (1)^2 + (2) = 4 + 1 + 2 = 7 \]

### Step 3: Calculate the Power of the Point
The power of the point (2, 1) with respect to the circle is given by:
\[ \text{Power} = S_1 - r^2 \]
where \(r\) is the radius of the circle. We found that \(r = \frac{1}{2}\), so:
\[ r^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]
Now, substituting \(S_1\) and \(r^2\) into the power formula:
\[ \text{Power} = 7 - \frac{1}{4} = \frac{28}{4} - \frac{1}{4} = \frac{27}{4} \]

### Final Answer
Thus, the power of the point (2, 1) with respect to the circle is:
\[ \frac{27}{4} \]

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