Find the Slope of the transverse common tangent to the circles x2 + y2 − 4x + 6y − 12 = 0 and x2 + y2 + 6x + 18y + 26 = 0
−512
Given circles,
x2+y2−4x−6y−12=0 - - - - - - (1)
x2+y2+6x+18y+26=0 - - - - - - (2)
Let the circle be c1 & c2 and radii r1 & r2 of circle (1) and (2) respectively.
c1(2, 3), c2(−3,−9)
r1=√g2+f2−c=√4+9+12=5
r2=√g2+f2−c=√9+81−26=8
c1c2=√(2+3)2+(3+9)2=√25+144=√169=13
c1c2=r1+r2
Both the circle touch each other externally.
Transverse common tangent is perpendicular to the line segment c1c2 and it divides line segment in the ratio of r1 and r2.
Product of slope of the two perpendicular line = -1
Slope of the line segment c1c2(m1)=y2−y1x2−x1=−9−3−3−2=−12−5=125
Slope of transverse common tangent =−1m1=−1125=−512
Correct option is A.