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Question: The centre of the circle \(x = 2 + 3 \cos \theta\), \(y = 3 \sin \theta - 1\) is....

The centre of the circle x=2+3cos⁡θx = 2 + 3 \cos \theta, y=3sin⁡θ−1y = 3 \sin \theta - 1 is.

A

(3, 3)

B

(2,−1)( 2 , - 1 )

C

(−2,1)( - 2,1 )

D

(−1,2)( - 1,2 )

Answer

(2,−1)( 2 , - 1 )

Explanation

Solution

x=2+3cos⁡θ,y=3sin⁡θ−1x = 2 + 3 \cos \theta , y = 3 \sin \theta - 1

x2+y2=4+9cos⁡2θ+12cos⁡θ+9sin⁡2θ+1−6sin⁡θx ^ { 2 } + y ^ { 2 } = 4 + 9 \cos ^ { 2 } \theta + 12 \cos \theta + 9 \sin ^ { 2 } \theta + 1 - 6 \sin \theta

=14+12cos⁡θ−6sin⁡θ= 14 + 12 \cos \theta - 6 \sin \theta =4(2+3cos⁡θ)−2(3sin⁡θ−1)+44 ( 2 + 3 \cos \theta ) - 2 ( 3 \sin \theta - 1 ) + 4

⇒ x2+y2=4x−2y+4x ^ { 2 } + y ^ { 2 } = 4 x - 2 y + 4

⇒ (x2−4x+4)+(y2+2y+1)=9\left( x ^ { 2 } - 4 x + 4 \right) + \left( y ^ { 2 } + 2 y + 1 \right) = 9

⇒ (x−2)2+(y+1)2=9( x - 2 ) ^ { 2 } + ( y + 1 ) ^ { 2 } = 9, ∴ centre is (2,−1)( 2 , - 1 ).