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A( 9,0), B(1,0) are two points. If P is a point such that PA:PB = 3:1, then the locus of P is

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$x^{2} + y^{2} = 9$

$x^{2} + y^{2} + 9 = 0$

$x^{2} - y^{2} = 9$

$x^{2} - y^{2} + 9 = 0$

answer is A.

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Detailed Solution

$\frac{PA}{PB} = \frac{3}{1} \Rightarrow {PA}^{2} = 9 {PB}^{2}$

$P A = 3 P B {(P A)}^{2} = 9 {(P B)}^{2} {(x - 9)}^{2} + {(y - 0)}^{2} = 9 [{(x - 1)}^{2} + {(y - 0)}^{2}] x^{2} + y^{2} = 9$

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