
Parametric equation of circle ${x^2} + {y^2} - 6x + 16y + 48 = 0$ are
Answer
553.2k+ views
Hint:
The parametric equation is those equations that can be written by using the third variable instead of using the same. So let suppose an equation ${x^2} + {y^2} + 2gx + 2fy + c = 0$ then the value for parametric equations are $x = - g + \sqrt {{g^2} + {f^2} - c} \cos \theta $ and $y = - f + \sqrt {{g^2} + {f^2} - c} \sin \theta $. On comparing the equation we can easily find the values of $g$ & $f$. Then put it into the parametric equation.
Formula used:
Let us suppose an equation of the represented as ${x^2} + {y^2} + 2gx + 2fy + c = 0$
The parametric equation for it will be given by
$x = - g + \sqrt {{g^2} + {f^2} - c} \cos \theta $, and $y = - f + \sqrt {{g^2} + {f^2} - c} \sin \theta $.
Complete step by step solution:
On comparing the equation ${x^2} + {y^2} - 6x + 16y + 48 = 0$ with the equation ${x^2} + {y^2} + 2gx + 2fy + c = 0$, we get
$ \Rightarrow - 6x = 2gx$
Now on canceling out the same term from both sides, we get
$ \Rightarrow g = - 3$
Similarly,
$ \Rightarrow 16y = 2fy$
Now on canceling out the same term from both sides, we get
$ \Rightarrow f = 8$
And also on comparing the equation we can get the value of $c$,
Therefore, $c = 48$.
Now by using the parametric equation formula,
$x = - g + \sqrt {{g^2} + {f^2} - c} \cos \theta $
On substituting the values, we get
$ \Rightarrow x = - \left( { - 3} \right) + \sqrt {{{\left( { - 3} \right)}^2} + {8^2} - 48} \cos \theta $
Now on solving the above values, we get
$ \Rightarrow x = 3 + \sqrt {19} \cos \theta $
And similarly, $y = - f + \sqrt {{g^2} + {f^2} - c} \sin \theta $
On substituting the values, we get
$ \Rightarrow y = - 8 + \sqrt {{{\left( { - 3} \right)}^2} + {8^2} - 48} \sin \theta $
Now on solving the above values, we get
$ \Rightarrow y = - 8 + \sqrt {19} \sin \theta $
Therefore, the parametric equations are $x = 3 + \sqrt {19} \cos \theta $ and $y = - 8 + \sqrt {19} \sin \theta $ respectively.
Note:
This type of question will look very confusing and tough but once you know the formula, it’s just become a matter of a few minutes to get through the answer. So the main thing in this question is to memorize the formula and then we can easily solve this type of problem. Sometimes in this type of question, you won’t have values given then that time we have to solve the equation, and then we will find the parametric equation.
The parametric equation is those equations that can be written by using the third variable instead of using the same. So let suppose an equation ${x^2} + {y^2} + 2gx + 2fy + c = 0$ then the value for parametric equations are $x = - g + \sqrt {{g^2} + {f^2} - c} \cos \theta $ and $y = - f + \sqrt {{g^2} + {f^2} - c} \sin \theta $. On comparing the equation we can easily find the values of $g$ & $f$. Then put it into the parametric equation.
Formula used:
Let us suppose an equation of the represented as ${x^2} + {y^2} + 2gx + 2fy + c = 0$
The parametric equation for it will be given by
$x = - g + \sqrt {{g^2} + {f^2} - c} \cos \theta $, and $y = - f + \sqrt {{g^2} + {f^2} - c} \sin \theta $.
Complete step by step solution:
On comparing the equation ${x^2} + {y^2} - 6x + 16y + 48 = 0$ with the equation ${x^2} + {y^2} + 2gx + 2fy + c = 0$, we get
$ \Rightarrow - 6x = 2gx$
Now on canceling out the same term from both sides, we get
$ \Rightarrow g = - 3$
Similarly,
$ \Rightarrow 16y = 2fy$
Now on canceling out the same term from both sides, we get
$ \Rightarrow f = 8$
And also on comparing the equation we can get the value of $c$,
Therefore, $c = 48$.
Now by using the parametric equation formula,
$x = - g + \sqrt {{g^2} + {f^2} - c} \cos \theta $
On substituting the values, we get
$ \Rightarrow x = - \left( { - 3} \right) + \sqrt {{{\left( { - 3} \right)}^2} + {8^2} - 48} \cos \theta $
Now on solving the above values, we get
$ \Rightarrow x = 3 + \sqrt {19} \cos \theta $
And similarly, $y = - f + \sqrt {{g^2} + {f^2} - c} \sin \theta $
On substituting the values, we get
$ \Rightarrow y = - 8 + \sqrt {{{\left( { - 3} \right)}^2} + {8^2} - 48} \sin \theta $
Now on solving the above values, we get
$ \Rightarrow y = - 8 + \sqrt {19} \sin \theta $
Therefore, the parametric equations are $x = 3 + \sqrt {19} \cos \theta $ and $y = - 8 + \sqrt {19} \sin \theta $ respectively.
Note:
This type of question will look very confusing and tough but once you know the formula, it’s just become a matter of a few minutes to get through the answer. So the main thing in this question is to memorize the formula and then we can easily solve this type of problem. Sometimes in this type of question, you won’t have values given then that time we have to solve the equation, and then we will find the parametric equation.
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