The locus of a point which is at a constant distance 5 from the fixed point $(2, 3)$ is:

x^{2} + y^{2} - 4 x - 6 y - 12 = 0

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x^{2} + y^{2} + 4 x - 6 y - 12 = 0

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x^{2} + y^{2} + 4 x + 6 y + 12 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x^{2} + y^{2} - 4 x - 6 y + 12 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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Solution

The correct option is A $x^{2} + y^{2} - 4 x - 6 y - 12 = 0$
Let $(x, y)$ be point then ATQ

$\sqrt{{(x - 2)}^{2} + {(y - 3)}^{2} = 5}$

${(x - 2)}^{2} + {(y - 3)}^{2} = 25$

$x^{2} + y^{2} + 4 + 9 - 4 x - 6 y = 25$

$\Rightarrow x^{2} + y^{2} - 4 x - 6 y - 12 = 0$

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