Equation of circle whose radius is 5 and which touches the circle x2+y2−2x−4y−20=0 at (5, 5) is
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
(x−9)2+(y−8)2=5
b
(x−9)2+(y+8)2=25
c
x2+y2=25
d
(x−9)2+(y−8)2=25
answer is D.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
C1=(1,2) r1=5 C2=(h,k) r2=5As the two circles radii are same the point where the two circles touch each other is the midpoint of the centres(5,5)=(h+12,k+22)⇒(h,k)=(9,8)∴(x−9)2+(y−8)2=25