Question

Obtain the parametric equation from the circle
$x^2+y^2-6x+4y-12=0$

Answer 1

Given equation of circle is $x^2+y^2-6x+4y-12=0$

$⟹x^2-6x+y^2+4y-12=0$

$⟹x^2-2\left(x\right)\left(3\right)+(3{)}^2-(3{)}^2+y^2+2\left(y\right)\left(2\right)+(2{)}^2-(2{)}^2-12=0$

$⟹(x-3{)}^2-9+(y+2{)}^2-4-12=0$

$⟹(x-3{)}^2+(y+2{)}^2=5^2$

Let $X=x-3,Y=y+2$

$⟹X^2+Y^2=5^2$

This is in the form of $x^2+y^2=a^2$ which as parametric equations as $x=a\cos \theta ,y=a\sin \theta$

$⟹X=5\cos \theta ,Y=5\sin \theta$

$⟹x-3=5\cos \theta ,y+2=5\sin \theta$

$⟹x=3+5\cos \theta ,y=-2+5\sin \theta$