Question

Please see the image and please answer this one

Answer 1

join the centres C1 and C2 with the respective points (say L and M) where the tangent touches the circles

radius is always perpendicular to tangent. thus C1L and C2M are perpendicular on tangent MN

now in the two triangles C1MP and C2NP

angle C1MP = angle C2NP = 90°

angle C1PM = angle C2PN ( vertically opposite angles)

angle PC1M = angle PC2N ( third angle)

thus

∆ C1MP ~ ∆ C2NP

C1P/C2P = C1M/C2N

C1P/ (35 - C1P) = 12/9 = 4/3. (C1C2 = 35)

3C1P = 4(35 - C1P)

3C1P = 140 - 4C1P

3C1P + 4C1P = 140

7C1P = 140

C1P = 140/7

C1P = 20 (option b)