Convert given circle into parametric form
i.${x^2} + {y^2} - 6x + 4y = 3$.
ii.${x^2} + {y^2} = 25$

Hint: We have formula for converting circle into parametric form i.e. The parametric coordinates of circle with the form ${(x - a)^2} + {(y - b)^2} = {r^2}$ is $(a + r\cos \theta ,b + r\sin \theta )$

Complete step-by-step answer:
Given equation of the circle is a
${x^2} + {y^2} - 6x + 4y = 3$
Make above equation perfect square by adding 9 in first equation and 4 in second equation
${x^2} - 6x + 9 + {y^2} + 4y + 4 = 3 + 9 + 4$
$\;{(x - 3)^2} + {(y + 2)^2} = {4^2}$
The parametric form will be
The parametric coordinates of circle with the form ${(x - a)^2} + {(y - b)^2} = {r^2}$is
$(a + r\cos \theta ,b + r\sin \theta )$ [θ being the parameter]
∴ The parametric coordinates of the given circle is $(3 + 4\cos \theta , - 2 + 4\sin \theta )$

ii)Since ${x^2} + {y^2} = 25$ is the equation of the circle centered at the origin with radius 5, its corresponding parametric equations are
$x\left( t \right) = 5cost$
$y\left( t \right) = 5sint,$
where $0 \leqslant t < 2\pi .$

Note: Remember formula of parametric equations of circles. Compare the given points with the standard equation of the circle and write the parametric equations.