The centre of the circle x - -1 - 2 -cos - y - 3 - 2 - sin - is -1- -3-1- 3-1- 3-None of these-

The correct option is B ( 1, 3)Given that x+1/2 = cos θ. Also y 3/2 = sin θ. ⇒ (x+1/2)^2 + (y 3/2)^2 = 1 ⇒ (x+1)^2 + (y 3)^2 = 4, whose centre is ( 1, 3). ...

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Standard XII

Mathematics

Definition of Circle

The centre of...

Question

The centre of the circle $x = - 1 + 2 c o s θ, y = 3 + 2 s i n θ,$ is

(1, -3)

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(-1, 3)

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(1, 3)

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None of these

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Solution

The correct option is B (-1, 3)
Given that $\frac{x + 1}{2} = c o s θ .$
Also $\frac{y - 3}{2} = s i n θ .$
$\Rightarrow {(\frac{x + 1}{2})}^{2} + {(\frac{y - 3}{2})}^{2} = 1 \Rightarrow (x + 1)^{2} + (y - 3)^{2} = 4,$
whose centre is (-1, 3).

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The centre of the circle x=−1+2 cos θ,y=3+2 sin θ, is

The correct option is B (-1, 3)Given that x+12=cos θ. Also y−32=sin θ. ⇒ (x+12)2+(y−32)2=1 ⇒(x+1)2+(y−3)2=4, whose centre is (-1, 3).

The centre of the circle $x = - 1 + 2 c o s θ, y = 3 + 2 s i n θ,$ is

The correct option is B (-1, 3)
Given that $\frac{x + 1}{2} = c o s θ .$
Also $\frac{y - 3}{2} = s i n θ .$
$\Rightarrow {(\frac{x + 1}{2})}^{2} + {(\frac{y - 3}{2})}^{2} = 1 \Rightarrow (x + 1)^{2} + (y - 3)^{2} = 4,$
whose centre is (-1, 3).