A point P moves so that length of tangent from P to the circle `x^(2)+y^(2)-2x-4y+1=0` is three times the distance of P from (1, - 2). Locus of P is

Options

A. straight line
B. circle
C. parabola
D. ellipse

Answer

To solve the problem, we need to find the locus of the point P that moves such that the length of the tangent from P to the given circle is three times the distance from P to the point (1, -2).

### Step-by-Step Solution:

1. **Identify the Circle Equation**:
The equation of the circle is given as:
\[ x^2 + y^2 - 2x - 4y + 1 = 0 \]
We can rewrite this in standard form by completing the square.

2. **Complete the Square**:
Rearranging the circle equation:
\[ (x^2 - 2x) + (y^2 - 4y) + 1 = 0 \]
Completing the square for \(x\) and \(y\):
\[ (x - 1)^2 - 1 + (y - 2)^2 - 4 + 1 = 0 \]
Simplifying gives:
\[ (x - 1)^2 + (y - 2)^2 = 4 \]
This shows that the center of the circle is (1, 2) and the radius is 2.

3. **Length of Tangent from Point P**:
The length of the tangent from a point \(P(h, k)\) to the circle is given by the formula:
\[ \text{Length of tangent} = \sqrt{(h - 1)^2 + (k - 2)^2 - 4} \]

4. **Distance from Point P to (1, -2)**:
The distance from point \(P(h, k)\) to the point (1, -2) is:
\[ \text{Distance} = \sqrt{(h - 1)^2 + (k + 2)^2} \]

5. **Set Up the Equation**:
According to the problem, the length of the tangent is three times the distance from P to (1, -2):
\[ \sqrt{(h - 1)^2 + (k - 2)^2 - 4} = 3 \sqrt{(h - 1)^2 + (k + 2)^2} \]

6. **Square Both Sides**:
Squaring both sides to eliminate the square roots:
\[ (h - 1)^2 + (k - 2)^2 - 4 = 9 \left((h - 1)^2 + (k + 2)^2\right) \]

7. **Expand and Simplify**:
Expanding both sides:
\[ (h - 1)^2 + (k - 2)^2 - 4 = 9\left((h - 1)^2 + (k^2 + 4k + 4)\right) \]
\[ (h - 1)^2 + (k - 2)^2 - 4 = 9(h - 1)^2 + 9k^2 + 36k + 36 \]
Rearranging gives:
\[ (h - 1)^2 + (k - 2)^2 - 9(h - 1)^2 - 9k^2 - 36k - 36 + 4 = 0 \]
Combine like terms:
\[ -8(h - 1)^2 - 8k^2 + 34k - 32 = 0 \]

8. **Divide by -8**:
Dividing through by -8:
\[ (h - 1)^2 + k^2 - \frac{34}{8}k + 4 = 0 \]

9. **Complete the Square for k**:
Completing the square for \(k\):
\[ (h - 1)^2 + \left(k - \frac{17}{8}\right)^2 = \frac{17^2}{64} - 4 \]
Simplifying gives:
\[ (h - 1)^2 + \left(k - \frac{17}{8}\right)^2 = \frac{17^2 - 256}{64} \]

10. **Final Locus Equation**:
The final equation represents the locus of point P. Replacing \(h\) with \(x\) and \(k\) with \(y\):
\[ (x - 1)^2 + \left(y - \frac{17}{8}\right)^2 = r^2 \]
where \(r^2\) is a constant derived from the above steps.