Question

,
1 Find the parametric equation of the Circle
x2 + y2 - 6x + 4y - 3 = 0 ​

Answer 1

Answer:

Given equation of circle is x

2

+y

2

−6x+4y−12=0

⟹x

2

−6x+y

2

+4y−12=0

⟹x

2

−2(x)(3)+(3)

2

−(3)

2

+y

2

+2(y)(2)+(2)

2

−(2)

2

−12=0

⟹(x−3)

2

−9+(y+2)

2

−4−12=0

⟹(x−3)

2

+(y+2)

2

=5

2

Let X=x−3,Y=y+2

⟹X

2

+Y

2

=5

2

This is in the form of x

2

+y

2

=a

2

which as parametric equations as x=acosθ,y=asinθ

⟹X=5cosθ,Y=5sinθ

⟹x−3=5cosθ,y+2=5sinθ

⟹x=3+5cosθ,y=−2+5sinθ

Answer 2

Answer:

Step-by-step explanation:

Given,

$\sf x^2+y^2-6x+4y-3=0$

To Find :-

Parametric Equation of the circle

Solution :-

$\sf x^2+y^2-6x+4y-3=0$

$\sf x^2-6x+y^2+4y-3=0$

$\sf x^2-2(x)(3)+(3)^2-(3)^2+y^2+2(y)(2)+(2)^2-(2)^2-3=0$

$\sf (x-3)^2+(y+2)^2-9-4-3=0$

$\sf (x-3)^2+(y-2)^2-16=0$

$\sf (x-3)^2+(y-2)^2=16$

$(x-3)^2+(y-2)^2=4^2$

Let ,

X = x - 3 , Y = y - 2

$\sf \implies X^2+Y^2=4^2$

This above equation in the form of " x^2 + y^2 = a^2" :-

Where :-

$\sf x=acos\theta,y=asin\theta$

$\sf \implies X = 4cos\theta , Y = 4sin\theta$

$x-3=4cos\theta,y-2=4sin\theta$

$x=4cos\theta+3,y=4sin\theta+2$