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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
i) AB,AP,AC are in A.P p) 0
ii) AB,AP,AC are in G.P q) $1 / \sqrt{2}$
iii) AB,AP, AC are in H.P r) $\sqrt{2}$
iv) AB, $\frac{A P}{2}$ , AC are in A.P s) $\sqrt{2} - 1$
t) 7
The correct option is

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$(i) \to (P), (i i) \to (Q), (i i i) \to (R), (i v) \to (S)$

$(i) \to (P), (i i) \to (S), (i i i) \to (Q), (i v) \to (R)$

$(i) \to (T), (i i) \to (S), (i i i) \to (R), (i v) \to (P)$

$(i) \to (P), (i i) \to (R), (i i i) \to (Q), (i v) \to (T)$

answer is B.

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Detailed Solution

$\begin{array}{l} x = - 1 + r \cos θ \\ y = 1 + r \sin θ \end{array}\} \Rightarrow$ the equation of circle ${(- 1 + r \cos θ)}^{2} + {(1 + r \sin θ)}^{2} = 1$

$\Rightarrow r^{2} - 2 r (\cos θ - \sin θ) + 1 = 0$

$A B = r_{1}, A C = r_{2}, A P = r$

a) Locus of p is ${(x + 1)}^{2} + {(y - 1)}^{2} = (x + 1) - (y - 1)$ it is a circle through the origin then $d = 0$

b ) $r^{2} = r_{1} r_{2} \Rightarrow {(x + 1)}^{2} + {(y - 1)}^{2} = 1$

$\Rightarrow d = \sqrt{2} - 1$

c) $r = \frac{2 r_{1} r_{2}}{r_{1} + r_{2}} \Rightarrow r \cos θ - r \sin θ = 1$

$\Rightarrow x - y + 1 = 0$

$\Rightarrow d = \frac{1}{\sqrt{2}}$

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

$\Rightarrow x^{2} + y^{2} = 2$

$\Rightarrow d = \sqrt{2}$

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