Let a point P be such that its distance from the point (5, 0) is thrice the distance of P from the point (-5, 0). If the locus of the point P is a circle of radius r, then 4r² is equal to ________. (Round off to the nearest integer)

Solution

Maths sir · 4 steps

Write the condition as PA² = 9 PB², expand it, and read off the circle's radius.

1. Set up the condition. Let P = (x, y), A = (5, 0) and B = (-5, 0). The condition PA = 3 PB has square roots in it, so square both sides to get PA² = 9 PB².

   (x-5)² + y² = 9[(x+5)² + y²]

2. Expand and tidy. This gives x² - 10x + 25 + y² = 9x² + 90x + 225 + 9y², so 8x² + 8y² + 100x + 200 = 0. Divide by 8 to get x² + y² + 25/2x + 25 = 0.
3. Read the radius. Compare with x² + y² + 2gx + 2fy + c = 0. Here g = 25/4, f = 0 and c = 25, so r² = g² + f² - c = 625/16 - 25 = 225/16.
4. So. 4r² = 4 × 225/16 = 225/4 = 56.25, which rounds to 56.

The common slip: 14. Stopping at r² = 225/16 = 14.06 and reporting it gives 14. The question asks for 4r², so multiply by 4.

The idea to keep. When the ratio of a point's distances from two fixed points is a constant other than 1, the locus is a circle. Square the distances, make the x² coefficient 1, then use r² = g² + f² - c.

Answer from the NTA final answer key. Solution by PrepWiser.