Match the elements of List I with elements of List II.
If $S = (x - 2)^{2} + (y + 1)^{2} = 1$ is a circle, then the equation of a:

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Solution

$S = {(x - 2)}^{2} + {(y + 1)}^{2} = 1$
Center is $(2, - 1)$ and radius is $r = 1$
A) Distance between x-axis and center is 1, hence $y = 0$ is equation of tangent
and distance from center to line $y + 2 = 0$ is $∣ ∣ ∣ ∣ \frac{- 1 + 2}{\sqrt{1^{2}}} ∣ ∣ ∣ ∣ = 1 = r$ , hence $y + 2 = 0$ is also tangent to the circle
B) $(2, - 1)$ satisfy $3 x + 4 y - 2 = 0 \Rightarrow 3 (2) + 4 (- 1) - 2 = 0 \Rightarrow 0 = 0$ and $x = 2 \Rightarrow 2 = 2$
hence $3 x + 4 y - 2 = 0$ and $x = 2$ is equation of diameter
C) Intersection point of tangent $y = 0$ and diameter $x = 2$ satisfy the equation of circle
and intersection point of tangent $y + 2 = 0$ and diameter $3 x + 4 y - 2 = 0$ satisfy the equation of circle hence, perpendicual line to tangent is $x = 2$ and $3 x + 4 y - 2$
D) For any circle diameter is the chord of maximum length i.e. $3 x + 4 y - 2 = 0$ and $x = 2$

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S \equiv x^{2} + y^{2} + x - y - 2 = 0

A, B, C, D

n \in I^{+}

List-I (Electronic configuration of elements )	List-II (I.E. in kJ/mol)
(a) $1 s^{2} 2 s^{2}$	(i) 801
(b) $1 s^{2} 2 s^{2} 2 p^{4}$	(ii) 899
(c) $1 s^{2} 2 s^{2} 2 p^{3}$	(iii) 1314
(d) $1 s^{2} 2 s^{2} 2 p^{1}$	(iv) 1402

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