AP EAPCET 08-Jul-2022 Shift 2 Paper

Let, c(h,k) be the centre of the given circle.

∵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.