The centre of the circle r2 - 2 - 4rcos- - 6 r sin- is

The correct option is B ( 2, 3) We have r ^ 2 =2 4rcosθ nbsp; +6rsinθ nbsp; Put x=rcosθ nbsp; and y=rsinθ nbsp; ⇒ x ^ 2 + y ^ ...

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Standard XII

Mathematics

Definition of Circle

The centre of...

Question

The centre of the circle $r^{2} = 2 - 4 r cos θ + 6 r sin θ$ is

(2, 3)

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(- 2, 3)

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(- 2, - 3)

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(2, - 3)

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Solution

The correct option is B $(- 2, 3)$
We have $r^{2} = 2 - 4 r cos θ + 6 r sin θ$
Put $x = r cos θ$ and $y = r sin θ$
$\Rightarrow x^{2} + y^{2} = 2 - 4 x + 6 y \Rightarrow x^{2} + y^{2} + 4 x - 6 y - 2 = 0$
$∴$ centre of the circle is $(- 2, 3)$

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The centre of the circle r2=2−4rcosθ+6rsinθ is

The correct option is B (−2,3)We have r2=2−4rcosθ+6rsinθPut x=rcosθ and y=rsinθ⇒x2+y2=2−4x+6y⇒x2+y2+4x−6y−2=0∴ centre of the circle is (−2,3)

The centre of the circle $r^{2} = 2 - 4 r cos θ + 6 r sin θ$ is

The correct option is B $(- 2, 3)$
We have $r^{2} = 2 - 4 r cos θ + 6 r sin θ$
Put $x = r cos θ$ and $y = r sin θ$
$\Rightarrow x^{2} + y^{2} = 2 - 4 x + 6 y \Rightarrow x^{2} + y^{2} + 4 x - 6 y - 2 = 0$
$∴$ centre of the circle is $(- 2, 3)$