Question

Two circles of radii a and b touch externally. If x is the radius of a third circle which is between them and touches them externally and also touching their direct common tangent, then 1/x is equal to

(a)  1/a + 1/b

(b)  /a + 1/b - 1/ab

(c)  1/a + 1/b + 2/√ab

(d)  1/a + 1/b + 1/√ab

Answer 1

Correct option (c)  1/a + 1/b + 2/√ab

Explanation :

Points A and B are the centres with radii a and b, respectively. LM is a direct common tangent. AP is drawn perpendicular to BM so that APML is a rectangle. Through the centre C of the circle with radius x, draw a line parallel to AP meeting the line AL at R and the line BM at Q. Now, Δ BCQ is right-angled triangle in which BC is the hypotenuse. We have BC = b + x and BQ = b - x. By Pythagoras theorem, we have

(BC)² =  (BQ)² +  (CQ)² ⇒ (b + x)² + (CQ)² 

√CQ = 2√(bx)  ....(1)

CR = 2√(ax)   ....(2)

Also AB  = b + a, BP = b - a and ∠APB =  90° which implies that 

Further AP= RQ = LM. Therefore, from Eqs. (1)– (3), we get