The correct options are
A The number of common tangents is 2.
C Length of the common tangent is (1500)1/4 units
Given circles are
x2+y2+2x+4y−20=0 and x2+y2+6x−8y+10=0
The centre and radius are,
C1=(−1,−2), r1=√1+4+20=5C2=(−3,4), r2=√9+16−10=√15
Now,
C1C2=√22+62=√40⇒r1+r2>C1C2>|r1−r2|
Therefore, the circles intersect in two distinct points, so they have 2 common tangents.
Now, the length of common tangent (DCT)
=√(d)2−(r1−r2)2=√40−(5−√15)2 (∵d=C1C2)=(1500)1/4