Consider two circles  $C_1$ of radius 'a' and $C_2$ of radius 'b' (b > a) both lying in the first  quadrant and touching the coordinate axes. In each of the conditions listed in column-I,  the ratio of  $\frac{b}{a}$ is given in column II.

	Column-I		Column-II
A)	$C_1$  and  $C_2$  touch each other	P)	$2+\sqrt{2}$
B)	$C_1$  and  $C_2$ are orthogonal	Q)	$3$
C)	$C_1$  and  $C_2$  intersect so that the common chord is longest	R)	$2+\sqrt{3}$
D)	$C_2$  passes through the centre of  $C_1$	S)	$3+2\sqrt{2}$

Equation of circle touching the coordinates axes and centre (r, r) in the first quadrant is
x² + y² – 2xr – 2yr + r² = 0
 
For r = a  or  b
hence C₁ : x² + y² – 2ax – 2ay + a² ....(1)   
 Centre (a, a),  radius = a, a > 0
C₂ : x² + y² – 2bx – 2by + b²  ....(2)   
 Centre (b, b),  radius b, b > 0
(A) C₁ and C₂ touch each other
radical axis between (1) and (2) is
(1) – (2) = 0
 2(b – a)x + 2(b – a)y – (b² – a²) = 0
 2x + 2y – (b + a) = 0 ....(3)
if it touches both C₁ and C₂ then perpendicular from (a, a) = radius 'a'
  $\left|\frac{2a+2a-(b+a)}{\sqrt{8}}\right|$  = a ....(4)
  | 3a – b | = $2\sqrt{2}a$  ....(5)
now origin and (a, a) must lie on the same side of (3)
but (0, 0) gives – ve sign with (3)
hence (a, a) should also give the same sign i.e.  4a – b – a < 0  $\Rightarrow$  3a – b < 0
Hence (5) becomes 
 b – 3a = $2\sqrt{2}a$     $\Rightarrow$     $\frac{b}{a}$ = $3+2\sqrt{2}$   Ans.  $\Rightarrow$  (S)
Alternatively: As  C₁ and C₂ touch each other externally so, 
 distance between their centre = sum of their radius
$\Rightarrow \sqrt{{(a-b)}^2+{(a-b)}^2}$    = (a + b)  $\Rightarrow$  2(a – b)² = (a + b)²    $\Rightarrow$ a² + b² – 6ab  = 0
$\Rightarrow$   $\frac{b^2}{a^2}-6\left(\frac{b}{a}\right)+1$ = 0
$\therefore$   $\frac{b}{a}$ = $\frac{6\pm \sqrt{36-4}}{2}$  =  $\frac{6\pm 4\sqrt{2}}{2}$  = $3\pm 2\sqrt{2}$