The parametric eqauation of the circle x 2-y 2-2 x-4 y-4-0 is -A- x-1-3 cos- y-2-3 sin-B- x-1-3 cos- y-2-3 sin-C- x-1-3 cos- y-2-3 sin-D- x-1-3 cos- y-2-3 sin-

The correct option is C x=1+3cos nbsp;θ, nbsp;y=2+3sin nbsp;θWe have, x2+y2 2x 4y 4=0⇒(x2 2x+1)+(y2 4y+4)=9⇒(x 1)2+(y 2)2=32 Hence the parametric form of ...

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Standard XII

Mathematics

Parametric Equation of a Circle

The parametri...

Question

The parametric eqauation of the circle $x^{2} + y^{2} - 2 x - 4 y - 4 = 0$ is .

x = 1 - 3 cos θ, y = 2 + 3 sin θ

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x = 1 + 3 cos θ, y = 2 - 3 sin θ

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x = 1 + 3 cos θ, y = 2 + 3 sin θ

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x = 1 - 3 cos θ, y = 2 - 3 sin θ

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Solution

The correct option is C $x = 1 + 3 cos θ, y = 2 + 3 sin θ$
We have,
$x^{2} + y^{2} - 2 x - 4 y - 4 = 0 \Rightarrow (x^{2} - 2 x + 1) + (y^{2} - 4 y + 4) = 9 \Rightarrow (x - 1)^{2} + (y - 2)^{2} = 3^{2}$
Hence the parametric form of the given circle will be :
$x = 1 + 3 cos θ, y = 2 + 3 sin θ$

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Similar questions

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x = 2 + 3 cos θ

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x = 3 c o s Θ, y = 3 s i n Θ, t h e n \frac{d y}{d x} = ?

Find the parametric equation of the circle $x^{2} + y^{2} - 2 x + 4 y - 11 = 0$

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The parametric eqauation of the circle x2+y2−2x−4y−4=0 is .

The correct option is C x=1+3cos θ, y=2+3sin θWe have, x2+y2−2x−4y−4=0⇒(x2−2x+1)+(y2−4y+4)=9⇒(x−1)2+(y−2)2=32 Hence the parametric form of the given circle will be : x=1+3cos θ, y=2+3sin θ

The parametric eqauation of the circle $x^{2} + y^{2} - 2 x - 4 y - 4 = 0$ is .

The correct option is C $x = 1 + 3 cos θ, y = 2 + 3 sin θ$
We have,
$x^{2} + y^{2} - 2 x - 4 y - 4 = 0 \Rightarrow (x^{2} - 2 x + 1) + (y^{2} - 4 y + 4) = 9 \Rightarrow (x - 1)^{2} + (y - 2)^{2} = 3^{2}$
Hence the parametric form of the given circle will be :
$x = 1 + 3 cos θ, y = 2 + 3 sin θ$