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

2 thoughts on “,<br />1 Find the parametric equation of the Circle<br />x2 + y2 – 6x + 4y – 3 = 0 ​”

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

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2. Answer:

   Step-by-step explanation:

   Given,

   [tex]\sf x^2+y^2-6x+4y-3=0[/tex]

   To Find :-

   Parametric Equation of the circle

   Solution :-

   [tex]\sf x^2+y^2-6x+4y-3=0[/tex]

   [tex]\sf x^2-6x+y^2+4y-3=0[/tex]

   [tex]\sf x^2-2(x)(3)+(3)^2-(3)^2+y^2+2(y)(2)+(2)^2-(2)^2-3=0[/tex]

   [tex]\sf (x-3)^2+(y+2)^2-9-4-3=0[/tex]

   [tex]\sf (x-3)^2+(y-2)^2-16=0[/tex]

   [tex]\sf (x-3)^2+(y-2)^2=16[/tex]

   [tex](x-3)^2+(y-2)^2=4^2[/tex]

   Let ,

   X = x – 3 , Y = y – 2

   [tex]\sf \implies X^2+Y^2=4^2[/tex]

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

   Where :-

   [tex]\sf x=acos\theta,y=asin\theta[/tex]

   [tex]\sf \implies X = 4cos\theta , Y = 4sin\theta[/tex]

   [tex]x-3=4cos\theta,y-2=4sin\theta[/tex]

   [tex]x=4cos\theta+3,y=4sin\theta+2[/tex]

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