Question

Parametric coordinates of any point of the circle \[{x^2} + {y^2} - 4x + 6y - 12 = 0\] are?

Answer 1

Hint: Here, we have to find the parametric co-ordinates of the circle. We have to find the radius by the equation given. Then by using the parametric equation of the circle we have to find the parametric co-ordinates. A parametric equation defines a group of quantities as functions of one or more independent variables called parameters.

Formula used:
We will use the following formulas:
The square of the sum of numbers is given by the algebraic identity \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
The square of the difference of numbers is given by the algebraic identity \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\]Parametric Equation of circle with centre (h, k) and radius R is given by
\[x = h + Rcos\theta ,\;y = k + Rsin\theta \], where \[\theta \] is the parameter.

Complete step-by-step answer:
We are given with the equation of circle \[{x^2} + {y^2} - 4x + 6y - 12 = 0\]
A quadratic equation of the circle is \[{x^2} + {y^2} + 2gx + 2fy + c = 0\]
Comparing the given equation with the quadratic equation, we have
\[2g = - 4\]; \[2f = 6\];
\[ \Rightarrow g = - 2;f = 3\]
Now, we have to convert the given equation to Cartesian form
\[ \Rightarrow \left( {{x^2} - 4x + 4} \right) + \left( {{y^2} + 6y + 9} \right) = 12 + 4 + 9\]
The square of the sum of numbers is given by the algebraic identity \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
The square of the Difference of numbers is given by the algebraic identity \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\]
\[ \Rightarrow {\left( {x - 2} \right)^2} + {\left( {y + 3} \right)^2} = 25\]
\[ \Rightarrow {\left( {x - 2} \right)^2} + {\left( {y + 3} \right)^2} = {5^2}\]
Now, the equation of the circle is of the form \[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\]
So, the centre of the circle is at (h, k)
Now, for this circle centre is at (2,-3) and radius is 5.
Parametric Equation of circle with centre (h, k) and radius R is given by
\[x = h + Rcos\theta ,\;y = k + Rsin\theta \] where \[\theta \] is the parameter.
Parametric Equation of circle with centre (2, -3) and radius 5, we have
\[ \Rightarrow x = 2 + 5\cos \theta ,y = - 3 + 5\sin \theta \]
Therefore, The Parametric co-ordinates of the circle are \[(2 + 5\cos \theta , - 3 + 5\sin \theta )\]


Note: We can find the radius of the circle using the formula \[\sqrt {{g^2} + {f^2} - c} \] .
\[ \Rightarrow \] Radius\[ = \sqrt {{2^2} + {3^2} - ( - 12)} = \sqrt {4 + 9 + 12} = \sqrt {25} = 5\]. Parametric equations are equations that depend on a single parameter. Equations can be converted between parametric equations and a single equation. Co-ordinate is the number representing the position of a line.