First slide

Cartesian plane

Question

$Given that A \equiv (1, - 1) and locus of B is x^{2} + y^{2} = 16 . If P divides A B in the ratio 3 : 2 then locus of P is$

Moderate

A

$(x - 2)^{2} + (y - 3)^{2} = 4$
B

$(x + 1)^{2} + (y - 2)^{2} = 4$
C

$(x - 3)^{2} + (y - 2)^{2} = 4$
D

$(3 x + 2)^{2} + (3 y - 2)^{2} = 400$

Solution

$Given point A (1, - 1) and B lies on x^{2} + y^{2} = 16 .$
$Let point B be (x, y) and point P be (h, k) .$
$\begin{aligned} ∴ (x, y) = (\frac{3 h + 2}{5}, \frac{3 k - 2}{5}) \\ Now B lies on x^{2} + y^{2} = 16 \end{aligned}$
$\begin{array}{l} ∴ {(\frac{3 h + 2}{5})}^{2} + {(\frac{3 k - 2}{5})}^{2} = 16 \\ ∴ (3 x + 2)^{2} + (3 y - 2)^{2} = 400, which is required locus. \end{array}$

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