The circles x2+y2+2g1x−a2=0 and x2+y2+2g2x−a2=0 cut each other orthogonally. If p1,p2 are perpendicular from (0,a) and (0,−a) on a common tangents of these circles then p1p2=
A
a2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
2a2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
3a2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
None of these
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is Aa2 Since the given circles cut each other orthogonally g1g2+a2=0 ...(1)
If lx+my=1 is a common tangent of these circles , then −lg1−1√l2+m2=±√g21+a2
⇒(lg1+1)2=(l2+m2)(g21+a2)
⇒m2g21−2lg1+a2(l2+m2)−1=0
Similarly, m2g22−2lg1+a2(l2+m2)−1=0
Thus, g1,g2 are the roots of the equation m2g2−2lg+a2(l2+m2)−1=0
⇒g1g2=a2(l2+m2)−1m2=−a2 by (1)
⇒a2(l2+m2)=1−a2m2 ...(2)
Now, p1p2=|ma−1|√l2+m2.|−ma−1|√l2+m2=∣∣1−m2a2∣∣l2+m2=1a2 by (2)