If the equation of two diameters of a circle are x+y=6

Let C (h, k) be the centre of the required circle. Since point of intersection of diameters is the centre of the circle,

x = h, y = k

∴ Equations of diameters become

2h + k = 6 …(i)

and 3h + 2k = 4 …(ii)

By (ii) – 2 x (i), we get

– h = – 8

∴ h = 8

Substituting h = 8 in (i), we get

2(8) + k = 6

∴ k = 6 – 16

∴ k = – 10

∴ Centre of the circle is C (8, –10) and radius, r = 9

The equation of a circle with centre at (h, k) and radius r is given by

(x – h)2 + (y – k)2 = r2

Here, h = 8, k = –10

∴ The required equation of the circle is

(x – 8)2 + (y +10)2 = 92

∴ x2 – 16x + 64 + y2 + 20y + 100 = 81

∴ x2 + y2 – 16x + 20y + 100 + 64 – 81 = 0

∴ x2 + y2 – 16x + 20y + 83 = 0.