Answer:
Option D
Explanation:
Let the centre of the circle be (h, k). Since the centre lies on the line y: x - 1
.'. k = h-1 ..(1)
Since the circle passes through the point (7, 3), therefore, the distance of the centre from this point is the radius of the circle.
.'. 3 = $\sqrt{(h-7)^{2}+(k-3)^{2}}$
3 = $\sqrt{(h-7)^{2}+(h-1-3)^{2}}$ using (1)
h = 7 or h = 4
F or h = 7, we get k = 6 and for h = 4, we get k = 3 Hence, the circles which satisfy the given conditions are:
$(x-7)^{2} +(y-6)^{2} =9$ or
$x^{2} + y^{2} -14x +12y +76 = 0$
and $(x -4)^{2} + (y- 3)^{2} = 9$ or
$x^{2} +y^{2} -8x-6y+16=0$
