Find the equation of a circle passing through the point (7, 3) having radius 3 units and whose centre lies on the line y = x – 1.
Since, the equation of a circle having centre (h, k), having radius as r units, is
(x – h)2 + (y – k)2 = r2Centre lies on the line i.e., y = x – 1,
Co – Ordinates are (h, k) = (h, h – 1)
(x – h)2 + (y – k)2 = r2
(7 – h)2 + (3 – (h – 1))2 = 32
49 + h2 – 14h + (3 – h +1)2 = 9
On rearranging we get
h2 – 14h + 49 +16 +h2 – 8h – 9 = 0
2h2 – 22h + 56 = 0
h2 – 11h + 28 = 0
h2 – 4h – 7h + 28 = 0
h (h – 4) – 7 (h – 4) = 0
(h – 7) (h – 4) = 0
h = 7 or 4
Centre = (7, 6) or (4, 3)
(x – h)2 + (y – k)2 = r2
Equation, having centre (7, 6)
(x – 7)2 + (y – 6)2 = 32
x2 – 14x + 49 + y2 – 12y + 36 – 9 = 0
x2 – 14x + y2 – 12y + 76 = 0
Equation, having centre (4, 3)
(x – 4)2 + (y – 3)2 = 32
x2 – 8x + 16 + y2 – 6y + 9 – 9 = 0
x2 – 8x + y2 – 6y + 16 = 0
Hence, the required equation is x2 – 14x + y2 – 12y + 76 = 0 or x2 – 8x + y2 – 6y + 16 = 0.