∵c(h,k) lies on the line x−y−1=0 ⇒h−k−1=0 ...(i) ⇒h=k+1 Now, CP is the radius. ⇒CP=3⇒(CP)2=9 ⇒(h−7)2+(k−3)2=9 ⇒(k−6)2+(k−3)2=9 ⇒2k2−18k+36=0 ⇒k2−9k+18=0⇒(k−6)(k−3)=0⇒k=6,3 ⇒h=6+1,3+1=7,4 Thus, (h,k)=(7,6) or (4,3). When C≡(7,6) and r=3 Equation of circle (x−7)2+(y−6)2=32 ⇒x2+y2−14x−12y+76=0 ∴ Option (d) is true.