Question

If a point P moves so that the distance from (0,2) to P is \(\frac{1}{√2 }\) times the distance of P from (-1,0), then the locus of the point P is

• A. a circle with centre (1, 4) and radius 10 units
• B. a circle with centre (-1, -4) and radius √10 units
• C. A circle with centre (1, 4) and radius √10 units
• D. a parabola with focus at (1,4) and length of latus rectum 10 units

Answer 1

The Correct Option is C

To find the locus of point $P = (x, y)$ such that the distance from $P$ to $(0, 2)$ is $\frac{1}{\sqrt{2}}$ times the distance from $P$ to $(-1, 0)$, we proceed as follows:

1. Expressing the Distances:
The distance from $P = (x, y)$ to $(0, 2)$ is:

$ \sqrt{(x-0)^2 + (y-2)^2} = \sqrt{x^2 + (y-2)^2} $
The distance from $P$ to $(-1, 0)$ is:

$ \sqrt{(x-(-1))^2 + (y-0)^2} = \sqrt{(x+1)^2 + y^2} $

2. Setting Up the Given Condition:
The problem states that the distance from $P$ to $(0, 2)$ is $\frac{1}{\sqrt{2}}$ times the distance from $P$ to $(-1, 0)$:

$ \sqrt{x^2 + (y-2)^2} = \frac{1}{\sqrt{2}} \sqrt{(x+1)^2 + y^2} $

3. Simplifying the Equation:
To eliminate the square roots, square both sides:

$ x^2 + (y-2)^2 = \frac{1}{2} ((x+1)^2 + y^2) $
Expand both sides:

Left side: $ (y-2)^2 = y^2 - 4y + 4 $, so:

$ x^2 + (y-2)^2 = x^2 + y^2 - 4y + 4 $
Right side: $ (x+1)^2 = x^2 + 2x + 1 $, so:

$ (x+1)^2 + y^2 = x^2 + 2x + 1 + y^2 $
Thus:

$ x^2 + y^2 - 4y + 4 = \frac{1}{2} (x^2 + 2x + 1 + y^2) $

4. Clearing the Fraction:
Multiply both sides by 2 to eliminate the fraction:

$ 2(x^2 + y^2 - 4y + 4) = x^2 + 2x + 1 + y^2 $
Expand the left side:

$ 2x^2 + 2y^2 - 8y + 8 = x^2 + 2x + 1 + y^2 $

5. Simplifying the Equation:
Move all terms to one side:

$ 2x^2 + 2y^2 - 8y + 8 - (x^2 + 2x + 1 + y^2) = 0 $
Combine like terms:

$ (2x^2 - x^2) + (2y^2 - y^2) - 8y - 2x + (8 - 1) = 0 $
$ x^2 + y^2 - 2x - 8y + 7 = 0 $

6. Completing the Square:
Rewrite the equation by completing the square for $x$ and $y$:

For $x$ terms: $ x^2 - 2x = (x^2 - 2x + 1) - 1 = (x - 1)^2 - 1 $
For $y$ terms: $ y^2 - 8y = (y^2 - 8y + 16) - 16 = (y - 4)^2 - 16 $
Substitute:

$ (x - 1)^2 - 1 + (y - 4)^2 - 16 + 7 = 0 $
Simplify:

$ (x - 1)^2 + (y - 4)^2 - 1 - 16 + 7 = 0 $
$ (x - 1)^2 + (y - 4)^2 - 10 = 0 $
$ (x - 1)^2 + (y - 4)^2 = 10 $

7. Interpreting the Result:
The equation $ (x - 1)^2 + (y - 4)^2 = 10 $ represents a circle with center $(1, 4)$ and radius $\sqrt{10}$.

Final Answer:
The locus of point $P$ is a circle with center $(1, 4)$ and radius $\sqrt{10}$ units.