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Question: Convert given circle into parametric form i.\({x^2} + {y^2} - 6x + 4y = 3\). ii.\({x^2} + {y^2} ...

Convert given circle into parametric form
i.x2+y2−6x+4y=3{x^2} + {y^2} - 6x + 4y = 3.
ii.x2+y2=25{x^2} + {y^2} = 25

Explanation

Solution

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=r2{(x - a)^2} + {(y - b)^2} = {r^2} is (a+rcos⁡θ,b+rsin⁡θ)(a + r\cos \theta ,b + r\sin \theta )

Complete step-by-step answer:
Given equation of the circle is a
x2+y2−6x+4y=3{x^2} + {y^2} - 6x + 4y = 3
Make above equation perfect square by adding 9 in first equation and 4 in second equation
x2−6x+9+y2+4y+4=3+9+4{x^2} - 6x + 9 + {y^2} + 4y + 4 = 3 + 9 + 4
  (x−3)2+(y+2)2=42\;{(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=r2{(x - a)^2} + {(y - b)^2} = {r^2}is
(a+rcos⁡θ,b+rsin⁡θ)(a + r\cos \theta ,b + r\sin \theta ) [θ being the parameter]
∴ The parametric coordinates of the given circle is (3+4cos⁡θ,−2+4sin⁡θ)(3 + 4\cos \theta , - 2 + 4\sin \theta )

ii)Since x2+y2=25{x^2} + {y^2} = 25 is the equation of the circle centered at the origin with radius 5, its corresponding parametric equations are
x(t)=5costx\left( t \right) = 5cost
y(t)=5sint,y\left( t \right) = 5sint,
where 0⩽t<2π.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.