What is the locus of a point which is at a distance $5$ units from $(- 2, 3)$ ?

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 an arbitrary point be $P (x, y)$
Then the distance of the point $P$ from the point $A (- 2, 3)$ is
$d = \sqrt{(x + 2)^{2} + (y - 3)^{2}}$
$d^{2} = (x + 2)^{2} + (y - 3)^{2}$
Now the distance is given to be $5$ ,
Hence
$(x + 2)^{2} + (y - 3)^{2} = 25$
$x^{2} + y^{2} + 4 x - 6 y + 4 + 9 = 25$
$x^{2} + y^{2} + 4 x - 6 y + 13 = 25$
$x^{2} + y^{2} + 4 x - 6 y - 12 = 0$ .

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