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Consider the 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. The value of b/a if
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 $3 + 2 \sqrt{2}$

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$A \to (s); B \to (r); C \to (q); D \to (p)$

$A \to (p, q); B \to (r); C \to (s); D \to (p)$

$A \to (r); B \to (q); C \to (p, s); D \to (q)$

$A \to (s); B \to (p, s); C \to (q); D \to (q)$

answer is D.

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Detailed Solution

$(A) C_{1} : x^{2} + y^{2} - 2 a (x + y) + a^{2} = 0$
Centre: $(a, a)$ , radius:a,
$C_{2}; x^{2} + y^{2} - 2 b (x + y) + b^{2} = 0$
Centre:(b,b), radius b
Since $C_{1} a n d C_{2}$ touch each other
$\Rightarrow \sqrt{2} (b - a) = a + b \Rightarrow \frac{b}{a} = {(\sqrt{2} + 1)}^{2} = 3 + 2 \sqrt{2}$
(B) $C_{1}$ and $C_{2}$ intersect orthogonally
$\Rightarrow 2 (a b + a b) = a^{2} + b^{2} \Rightarrow b^{2} - 4 a b + a^{2} = 0 \Rightarrow \frac{b}{a} = 2 + \sqrt{3}$
(C) The common chord is
$2 (b - a) (x + y) = b^{2} - a^{2}$ .
(D) $C_{2}$ passes through (a,a)
$\Rightarrow 2 a^{2} - 4 a b + b^{2} = 0 \Rightarrow \frac{b}{a} = 2 + \sqrt{2}$

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