Find the equation of a circle passing through

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:

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.

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