Question

obtain the parametric equation of the circle x square + y square -6x+4y-12 = 0

Answer 1

SOLUTION

TO DETERMINE

The parametric equation of the circle

$\sf{ {x}^{2} + {y}^{2} - 6x + 4y - 12 = 0}$

EVALUATION

Here the given equation of the circle is

$\sf{ {x}^{2} + {y}^{2} - 6x + 4y - 12 = 0}$

Now we rewrite it as below

$\sf{ {x}^{2} + {y}^{2} - 6x + 4y - 12 = 0}$

$\sf{ \implies{x}^{2} - 6x + {y}^{2} + 4y - 12 = 0}$

$\sf{ \implies{x}^{2} - 6x + 9+ {y}^{2} + 4y + 4- 25 = 0}$

$\sf{ \implies{(x - 3)}^{2} + {(y + 2)}^{2} = 25 }$

$\sf{ \implies{(x - 3)}^{2} + {(y + 2)}^{2} = {5}^{2} }$

$\displaystyle \sf{ \implies \: \frac{ {(x - 3)}^{2} }{ {5}^{2} } + \frac{ {(y + 2)}^{2} }{ {5}^{2} } = 1 }$

$\displaystyle \sf{ \implies \: { \bigg( \frac{x - 3}{5} \bigg)}^{2} + { \bigg( \frac{y + 2}{5} \bigg)}^{2} = 1 }$

Take

$\displaystyle \sf{ \implies \: \frac{x - 3}{5} = \cos \theta \: \: and \: \: \: \frac{y + 2}{5} = \sin \theta }$

$\displaystyle \sf{ \implies \: x = 3 + 5 \cos \theta \: \: and \: \: \: y = - 2 + 5\sin \theta }$

FINAL ANSWER

Hence the required parametric equation of the circle is

$\displaystyle \sf{ \: x = 3 + 5 \cos \theta \: \: and \: \: \: y = - 2 + 5\sin \theta }$

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