Given circles, C1:x2+y2+30x−2y+1=0 ⇒centre(O)=(−15,1) ⇒radius(r1)=√225+1−1=15 And, C2:x2+y2−14x+6y+33=0 ⇒centre(O′)=(−15,1) ⇒ radius (r2)=√49+9−33=5 since point T divides OO′ in the ratio of 15: 5 i.e., 3: 1 internally. The co-ordinates of point T is T=(
21−15
4
,
−9+1
4
) T=(
3
2
,−2) since point D divides OO′ in the ratio of 15: 5 i.e., 3: 1 externally. The co-ordinates of point D is, D=(
21+15
4
,
−9−1
4
) D=(18,−5) The centre of circle with TD as diameter is C=(