A point P moves in such a way that the ratio | vedclass

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(B) Let the two fixed coplanar points be $A(0, 0)$ and $B(a, 0)$.Let the moving point $P$ be $(x, y)$.According to the given condition,the ratio of di

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Circle

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(B) Let the two fixed coplanar points be $A(0, 0)$ and $B(a, 0)$.Let the moving point $P$ be $(x, y)$.According to the given condition,the ratio of distances is constant,say $\lambda$:$\frac{PA}{PB} = \lambda \implies \frac{\sqrt{x^2 + y^2}}{\sqrt{(x-a)^2 + y^2}} = \lambda$Squaring both sides:$x^2 + y^2 = \lambda^2 ((x-a)^2 + y^2)$$x^2 + y^2 = \lambda^2 (x^2 - 2ax + a^2 + y^2)$$(1 - \lambda^2)x^2 + (1 - \lambda^2)y^2 + 2a\lambda^2 x - a^2\lambda^2 = 0$Since $\lambda 
e 1$,we can divide by $(1 - \lambda^2)$:$x^2 + y^2 + \frac{2a\lambda^2}{1 - \lambda^2}x - \frac{a^2\lambda^2}{1 - \lambda^2} = 0$This is the general equation of a circle $x^2 + y^2 + 2gx + 2fy + c = 0$. Thus,the locus is a circle.

$A$ point $P$ moves in such a way that the ratio of its distance from two coplanar points is always a fixed number $(\lambda \ne 1)$. Then its locus is

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