A variable straight line through $A (- 1, 1)$ is drawn to cut the circle $x^{2} + y^{2} = 1$ at the points B and C. A point ‘P’ is chosen on the line ABC satisfying the condition given the Column - I. Let d be the minimum distance of the origin from the locus of P given in the Column - II
COLUMN - I COLUMN - II
(A) AB, AP, AC are in A.P (p) 0
(B) AB, AP, AC are in G.P (q) $1 / \sqrt{2}$
(C) AB, AP, AC are in H.P (r) $\sqrt{2}$
(D) AB, $\frac{AP}{2},$ AC are in A.P. (s) $\sqrt{2} - 1$
A B C D
1) p s q r
2) r q s p
3) q r p s
4) s p q r

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Detailed Solution

$x = - 1 + rcos θ, y = 1 + rsin θ,$ the equation of the circle $(- 1 + rcos θ) 2 + (1 + rsin θ) 2 = 1$

$\begin{array}{l} r^{2} - 2 r (\cos θ - \sin θ) + 1 = 0 \\ \Rightarrow If AB = r_{1}, AC = r_{2}, AP = r \end{array}$

A) $2 A P = A B + A C \Rightarrow 2 r = r_{1} + r_{2}$

$2 r = 2 (\cos θ - \sin θ) \Rightarrow r^{2} = r \cos θ - r \sin θ$

locus of P is ${(x + 1)}^{2} + {(y - 1)}^{2} = (x + 1) - (y - 1) \Rightarrow x^{2} + y^{2} + x - y = 0$

It is a circle through the origin then d = 0

B) $r^{2} = r_{1} r_{2} \Rightarrow {(x + 1)}^{2} + {(y - 1)}^{2} = 1 \Rightarrow d = O C - r = \sqrt{2} - 1$

C) $r = \frac{2 r_{1} r_{2}}{r_{1} + r_{2}} \Rightarrow rcos θ - rsin θ = 1$

$\Rightarrow x + 1 - y + 1 = 1 x - y + 1 = 0 d = \frac{1}{\sqrt{2}}$

$\begin{array}{l} D) r = r_{1} + r_{2} \Rightarrow (x + 1)^{2} + (y - 1)^{2} = 2 (x + 1 - y + 1) \\ \Rightarrow x^{2} + y^{2} = 2 \\ \Rightarrow d = \sqrt{2} \end{array}$

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COLUMN - I		COLUMN - II
(A)	AB, AP, AC are in A.P	(p)	0
(B)	AB, AP, AC are in G.P	(q)	$1 / \sqrt{2}$
(C)	AB, AP, AC are in H.P	(r)	$\sqrt{2}$
(D)	AB, $\frac{AP}{2},$ AC are in A.P.	(s)	$\sqrt{2} - 1$