Vikas TU
Last Activity: 7 Years ago
Dear Student,
let centres of circles be A and B . AB = 5
Extend AB to C to intersect direct common tangent PQ at C
By similar triangles CQB and CPA,
BC/(BC+5) = 2/3
Similarly, QC/(QC+PQ) = 2/3
We know that QC = √96 by applying Pythagoras th in triangle BQC
Therefore length of Direct common tangent PQ = 2√6 .
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)