Question

Find the equation circle if the equations of two diameters are 2x + y = 6 and 3x + 2y = 4. When radius of circle is 9

Answer 1

Given equations of diameters are 2x + y = 6 and 3x + 2y = 4.

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.