If D is the diameter of the larger circle and d is the diameter of the small circle then D+d=?
(A) a+b (B) 2(a+b) (C) √2(a+b) (D) 2√ab​

NOTE: I'm assuming that there is a right angled triangle whose sides are tangents to the smaller circle.
Ans is a+b
Solution: Let the triangle be P
                                              |       
                                              y        x
                                              |              
                                             Q----z------ R

where P,Q and R are vertices and x, y and z are points of intersection with inner circle.

Now, it is clear that sides of triangle are tangent to the inner circle.
So, as tangents from an external point are equal,

• PY = PX
• QY = QZ
• RX = RZ

​​By this, we have PX + RX = PY + RZ = D
In smaller cicle, let the centre be O
As tangents are perpendicular to radius,

• OY ​| QY
• OZ ​| QZ

OYQZ forms a square, 2OY = QY + QZ = d

D + d = PY + RZ + ​QY + QZ = PQ + QR
Hence, D + d = a + b