Question

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

Answer 1

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.