RS is the chord of contact of the point P on a circle center at O. if $m_{1} = s l o p e o f ¯ ¯¯¯¯¯¯ ¯ R S a n d m_{2} = s l o p e o f ¯ ¯¯¯¯¯¯ ¯ O P$

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Solution

The correct option is D

The circle described in the question is as given

It can be observed that $Δ P N R a n d Δ P N S$ are congruent triangle since length of the tangents are equal. This implies $∠ R N P = ∠ P N S = 90^{\circ}$

$\Rightarrow ¯ ¯¯¯¯¯¯ ¯ P O ⊥ ¯ ¯¯¯¯¯¯ ¯ R S$

$∴$ Product of the slopes of $⊥$ r lines is -1

Hence $m_{1} . m_{2} = - 1$

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RS is the chord of contact of the point P on a circle center at O. if m1=slope of ¯¯¯¯¯¯¯¯RS and m2=slope of ¯¯¯¯¯¯¯¯OP

RS is the chord of contact of the point P on a circle center at O. if $m_{1} = s l o p e o f ¯ ¯¯¯¯¯¯ ¯ R S a n d m_{2} = s l o p e o f ¯ ¯¯¯¯¯¯ ¯ O P$