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.