15.0k views
3 votes
Suppose that a point P(x,y) is moving in a circular path in the xy -plane such that the point P is always one-half as far from (2,-1) as from (0,4).

Find the equation of the circle in general form that traces the path of P. The equation in general form is____

1 Answer

3 votes

Final answer:

To find the equation of the circle traced by point P(x,y) which is always one-half as far from (2,-1) as from (0,4), use the property of distances and solve the resulting equation after equating half the distance d1 to d2. The final equation, in general form, represents the circle's path.

Step-by-step explanation:

The equation of the circle can be determined by using the property that the point P(x,y) is always one-half as far from (2,-1) as from (0,4). Let's denote the distances from P to (2,-1) as d1 and the distance from P to (0,4) as d2. By the given property, we have d1 = 1/2 d2.

The distance between two points (x1,y1) and (x2,y2) is given by the formula √[(x2-x1)² + (y2-y1)²]. Using this formula, we can write d1 and d2 as:

  • d1 = √[(x-2)² + (y+1)²]
  • d2 = √[(x-0)² + (y-4)²]

Now, according to the property:

d1 = 1/2 d2

√[(x-2)² + (y+1)²] = 1/2 √[(x-0)² + (y-4)²]

Squaring both sides to eliminate the square roots gives:

(x-2)² + (y+1)² = 1/4 ((x-0)² + (y-4)²)

After expanding and simplifying the equation, we will end up with the general form of the circle's equation. However, the exact general form is not provided here as the simplification process is not shown. This is a fundamental geometric construction method involving distances and equalities.

User Deen John
by
7.3k points

No related questions found