Parametric coordinates of any point of the circle $x^{2} + y^{2} - 4 x + 6 y - 12 = 0$ are?

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Solution

Given equation of the circle is
$x^{2} + y^{2} - 4 x + 6 y - 12 = 0$
$\Rightarrow x^{2} - 4 x + 4 + y^{2} + 6 y + 9 = 12 + 4 + 9$
$\Rightarrow (x - 2)^{2} + (y + 3)^{2} = 5^{2}$
The parametric coordinates of circle with the form $(x - a)^{2} + (y - b)^{2} = r^{2}$ is
$(a + r cos θ, b + r sin θ)$ [ $θ$ being the parameter]
$∴$ The parametric coordinates of the given circle is
$(2 + 5 cos θ, - 3 + 5 sin θ)$

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