Given x2+y2−4x−6y−12=0
C1(2,3),r1=22+32+12=25=5
And x2+y2+6x+18y+26=0
C2(−3,−9),r2=32+92−26
=90−26=8
Then, C1C2=52+122=13
C1C2=r1+r2
⇒Externally touching circles.
⇒3 common tangents.
JEE Main 2015 — Mathematics Coordinate Geometry
The number of common tangents to the circles x2+y2−4x−6y−12=0 and x2+y2+6x+18y+26=0, is
Held on 4 Apr 2015 · Verified 6 Jul 2026.
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