(b,c,d)Equation of a curve passing through the intersection points of the given curves

ax 2 + 2hxy + by 2 – 2gx – 2fy + c = 0 ........(1)

and a ′ x 2 – 2hxy + (a ′ + a – y 2 – 2g ′ x – 2f ′ y + c = 0 ........(2)

can be written as {a ′ x 2 – 2hxy +(a ′ + a – y 2 – 2g ′ x – 2f ′ y + c}

+ λ {ax 2 + 2hxy + by 2 – 2gx – 2fy + c} = 0

i.e. (a ′ + λ a)x 2 + 2h( λ – 1)xy + (a ′ + a – b + λ y 2

– 2(g ′ + λ g)x – 2(f ′ + λ f)y + (1 + λ ) c = 0 ........(3)

According to the given condition equation (3) must represent a circle, therefore, we have

coeff. of x 2 = coeff. of y 2

i.e. a ′ + λ a = a ′ + a – b + λ b i.e. λ (a – = a – b

gives λ = 1 and coeff. of xy = 0 i.e. λ – 1 = 0 gives λ = 1.

The identical values prove that the curve is a circle.

Putting the above value of λ in equation (3) gives the equation of the circle passing through the intersection points of the curves represented by equations (1) and (2) as (a ′ +a)(x 2 + y 2 ) – 2
(g ′ + g)x – 2(f ′ + f)y + 2c = 0

which has its centre at the point

We can see that the coordinates of the given point P is the same as the centre of the circle passing through the points A, B, C and D. Therefore, we have PA 2 = PB 2 = PC 2 = PD 2 = radius of the circle which gives the desired result PA 2 + PB 2 + PC 2 = 3PD 2 .