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Question: The centre of the circle r<sup>2</sup> = 2 – 4r cos θ + 6r sin θ is...

The centre of the circle r2 = 2 – 4r cos θ + 6r sin θ is

A

(2, 3)

B

(–2, 3)

C

(–2, –3)

D

(2, –3)

Answer

(–2, 3)

Explanation

Solution

Put x = r cos θ & y = r sin θ ⇒ x2 + y2 = 2 –4x + 6y

⇒ x2+y2 + 4x –6y – 2 = 0; ∴ Centre = (–2, 3)