If P , Q are conjugate points with respect to a circle S ≡ x 2 + y 2 + 2 g x + 2 f y + c = 0 , then prove that the circle P Q as diameter cuts the circle S = 0 orthogonally.

Given circle: S ≡ x 2 + y 2 + 2 g x + 2 f y + c = 0 Let P x 1 , y 1 , Q x 2 , y 2 be the conjugate Points w.r.t. the circle S = 0 Since, P x 1 , y 1 , Q x 2 , y 2 are conjugate Points, then S 12 = 0 ⇒ x 1 x 2 + y 1 y 2 + g x 1 + x 2 + f y 1 + y 2 + c = 0 ⇒ x 1 x 2 + y 1 y 2 + c = - g x 1 + x 2 - f y 1 + y 2 → 1 Equation of the circle on P Q as diameter is S ' = x - x 1 x - x 2 + y - y 1 y - y 2 = 0 ⇒ S ' = x 2 + y 2 - x 1 + x 2 x - y 1 + y 2 y + x 1 x 2 + y 1 y 2 = 0 And also, given that, S = 0 and S ' = 0 are orthogonal, then 2 g 1 g 2 + 2 f 1 f 2 = 2 g - x 1 + x 2 2 + 2 f - y 1 + y 2 2 = - g x 1 + x 2 - f y 1 + y 2 = x 1 x 2 + y 1 y 2 + c {Using 1 } and C + c ' = x 1 x 2 + y 1 y 2 + c ∴ 2 g 1 g 2 + 2 f 1 f 2 = C + c ' Hence, circles are orthogonal to each other.

If P , Q are conjugate points with respect to a circle S ≡ x 2 + y 2 + 2 g x + 2 f y + c = 0 , then prove that the circle P Q as diameter cuts the circle S = 0 orthogonally.

Given circle: S ≡ x 2 + y 2 + 2 g x + 2 f y + c = 0 Let P x 1 , y 1 , Q x 2 , y 2 be the conjugate Points w.r.t. the circle S = 0 Since, P x 1 , y 1 , Q x 2 , y 2 are conjugate Points, then S 12 = 0 ⇒ x 1 x 2 + y 1 y 2 + g x 1 + x 2 + f y 1 + y 2 + c = 0 ⇒ x 1 x 2 + y 1 y 2 + c = - g x 1 + x 2 - f y 1 + y 2 → 1 Equation of the circle on P Q as diameter is S ' = x - x 1 x - x 2 + y - y 1 y - y 2 = 0 ⇒ S ' = x 2 + y 2 - x 1 + x 2 x - y 1 + y 2 y + x 1 x 2 + y 1 y 2 = 0 And also, given that, S = 0 and S ' = 0 are orthogonal, then 2 g 1 g 2 + 2 f 1 f 2 = 2 g - x 1 + x 2 2 + 2 f - y 1 + y 2 2 = - g x 1 + x 2 - f y 1 + y 2 = x 1 x 2 + y 1 y 2 + c {Using 1 } and C + c ' = x 1 x 2 + y 1 y 2 + c ∴ 2 g 1 g 2 + 2 f 1 f 2 = C + c ' Hence, circles are orthogonal to each other.