Without changing the direction of the axes, the origin is transferred to the point $(2, 3)$ . Then the equation $x^{2} + y^{2} - 4 x - 6 y + 9 = 0$ changes to

x^{2} + y^{2} + 4 = 0

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x^{2} + y^{2} = 4

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x^{2} + y^{2} - 8 x - 12 y + 48 = 0

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

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Solution

The correct option is B $x^{2} + y^{2} = 4$
Given equation
$x^{2} + y^{2} - 4 x - 6 x + 9 = 0 \Rightarrow x^{2} - 4 x + 2^{2} + y^{2} - 6 x + 3^{2} = 2^{2} \Rightarrow (x - 2)^{2} + (y - 3)^{2} = 4$
$∵ x - 2 \to x, y - 3 \to y ∴ the new equation will be x^{2} + y^{2} = 4$

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(2, 3)

x^{2} + y^{2} - 4 x - 6 y + 9 = 0

A B

A (0, 3)

x^{2} + 4 x + (y - 3)^{2} = 0,

M

A M = 2 A B

M

x^{2} + y^{2} + 8 x + 10 y - 7 = 0

x^{2} + y^{2} - 4 x - 6 y = 0

x^{2} + 4 x + (y - 3)^{2} = 0

x^{2} + y^{2} - 8 x + 6 y - 5 = 0

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