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Question
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.
Solution
Let the equation of the circle be (x – h)2 + (y – k)2 = r2
If it passes through (7, 3)
Then (7 – h)2 + (3 – k)2 = (3)2 ......[∵ r = 3]
⇒ 49 + h2 – 14h + 9 + k2 – 6k = 9
⇒ h2 + k2 – 14h – 6k + 49 = 0 ......(i)
If centre (h, k) lies on the line y = x – 1 then
k = h – 1 .......(ii)
Putting the value of k in equation (i) we get
h2 + (h – 1)2 – 14h – 6(h – 1) + 49 = 0
⇒ h2 + h2 + 1 – 2h – 14h – 6h + 6 + 49 = 0
⇒ 2h2 – 22h + 56 = 0
⇒ h2 – 11h + 28 = 0
⇒ h2 – 7h – 4h + 28 = 0
⇒ h(h – 7) – 4(h – 7) = 0
⇒ (h – 4)(h – 7) = 0
∴ h = 4, h = 7
From equation (ii) we get
k = 4 – 1 = 3 and k = 7 – 1 = 6.
So, the centres are (4, 3) and (7, 6).
∴ Equation of the circle is
Taking centre (4, 3),
(x – 4)2 + (y – 3)2 = 9
x2 + 16 – 8x + y2 + 9 – 6y = 9
⇒ x2 + y2 – 8x – 6y + 16 = 0
Taking centre (7, 6)
(x – 7)2 + (y – 6)2 = 9
⇒ x2 + 49 – 14x + y2 + 36 – 12y = 9
⇒ x2 + y2 – 14x – 12y + 76 = 0
Hence, the required equations are x2 + y2 – 8x – 6y + 16 = 0 and x2 + y2 – 14x – 12y + 76 = 0.
