Question

Polar equations of the circle for the given coordinate (x,y) which satisfies the equation given by (x-a)^2+(y-b)^2=r^2 where (a,b) is the coordinates of the centre of the circle &r is the radius.

(a) x = r cos⁡θ, y = r sin⁡θ

(b) x = a+ r cos⁡θ, y = b + r sin⁡θ

(c) y = a+r cos⁡θ, x = b + r sin⁡θ

(d) x = r sin⁡θ, y = r cos⁡θ

I had been asked this question in semester exam.

This intriguing question comes from Polar Curves in division Differential Calculus of Engineering Mathematics

Answer 1

Right answer is (b) x = a+ r cos⁡θ, y = b + r sin⁡θ

For explanation I would say: option x = a+ r cos⁡θ, y = b + r sin⁡θ satisfies the equation (x-a)^2+(y-b)^2=r^2

because LHS=(a+r cos⁡θ-a)^2 + (b+ r sin⁡θ-b)^2 = r^2(cos^2 θ + sin^2 θ)= r^2=RHS.