We need to find equation of a circle whose `r = 3`.
Center lies on the line `y = x - 1`
`:.` Coordinates of center: `C(x_1, x_1-1)`
General equation of the circle:
`(x-h)^2 + (y-k)^2 = r^2`
Let cordinates of center be `(x_1,x_1-1)`
`rArr (x-x_1)^2 + (y - (x_1-1))^2 = 9`
Given point `(7,3)` passes through the circle.
`rArr (7-x_1)^2 + (3 - x_1 +1)^2 = 9`
`rArr 49 + x_1^2 - 14x_1 + 16 + x_1^2 - 8x_1 = 9`
`rArr 2x_1^2 - 22x_1 + 56 = 0`
` rArr x_1^2 - 11x_1 + 28 = 0`
`rArr x_1^2 - 7x_1 - 4x_1 + 28 = 0`
`rArr x_1(x_1 - 7) -4(x_1 - 7) = 0`
`x_1 = 4 or 7`
This means center coordinates are `C(4,3)` or `C(7,6)`
`:.` Equations of circle:
`(x-4)^2 + (y-3)^2 = 9`
or
`(x-7)^2 + (y-6)^2 = 9`
