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(D) Let $C(h, k)$ be the centre of the given circle.Since $C(h, k)$ lies on the line $x-y-1=0$,we have $h-k-1=0$,which implies $h=k+1$.Given that the

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$x^2+y^2-14x-12y+76=0$

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(D) Let $C(h, k)$ be the centre of the given circle.Since $C(h, k)$ lies on the line $x-y-1=0$,we have $h-k-1=0$,which implies $h=k+1$.Given that the radius is $3$,the distance from $C(h, k)$ to $P(7, 3)$ is $3$.So,$(h-7)^2 + (k-3)^2 = 3^2 = 9$.Substituting $h=k+1$,we get $(k+1-7)^2 + (k-3)^2 = 9$.$(k-6)^2 + (k-3)^2 = 9$.$k^2 - 12k + 36 + k^2 - 6k + 9 = 9$.$2k^2 - 18k + 36 = 0$.$k^2 - 9k + 18 = 0$.$(k-6)(k-3) = 0$,so $k=6$ or $k=3$.If $k=6$,then $h=7$,so $C=(7, 6)$. The equation is $(x-7)^2 + (y-6)^2 = 9$,which simplifies to $x^2 - 14x + 49 + y^2 - 12y + 36 = 9$,or $x^2 + y^2 - 14x - 12y + 76 = 0$.If $k=3$,then $h=4$,so $C=(4, 3)$. The equation is $(x-4)^2 + (y-3)^2 = 9$,which simplifies to $x^2 - 8x + 16 + y^2 - 6y + 9 = 9$,or $x^2 + y^2 - 8x - 6y + 16 = 0$.Comparing with the given options,$x^2 + y^2 - 14x - 12y + 76 = 0$ is present as option $(d)$.

If a circle of radius $3$ passes through the point $(7,3)$ and has its centre on the line $x-y-1=0$,then its equation among the following is

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