Question:
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 ________
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 ________
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The ratio of distances from two fixed points being constant ($k \neq 1$) results in a {Circle of Apollonius}.
Updated On: Feb 9, 2026
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Correct Answer: 56
Solution and Explanation
Step 1: Let $P = (x, y)$, $A = (5, 0)$, $B = (-5, 0)$. Given $PA = 3PB$.
Step 2: $(x-5)^2 + y^2 = 9[(x+5)^2 + y^2]$.
Step 3: $x^2 - 10x + 25 + y^2 = 9[x^2 + 10x + 25 + y^2]$.
Step 4: $8x^2 + 8y^2 + 100x + 200 = 0 \Rightarrow x^2 + y^2 + \frac{25}{2}x + 25 = 0$.
Step 5: $r^2 = (\frac{25}{4})^2 - 25 = \frac{625}{16} - \frac{400}{16} = \frac{225}{16}$.
Step 6: $4r^2 = 4 \times \frac{225}{16} = \frac{225}{4} = 56.25$.
Step 2: $(x-5)^2 + y^2 = 9[(x+5)^2 + y^2]$.
Step 3: $x^2 - 10x + 25 + y^2 = 9[x^2 + 10x + 25 + y^2]$.
Step 4: $8x^2 + 8y^2 + 100x + 200 = 0 \Rightarrow x^2 + y^2 + \frac{25}{2}x + 25 = 0$.
Step 5: $r^2 = (\frac{25}{4})^2 - 25 = \frac{625}{16} - \frac{400}{16} = \frac{225}{16}$.
Step 6: $4r^2 = 4 \times \frac{225}{16} = \frac{225}{4} = 56.25$.
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