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

$m_{1} + m_{2} = 0$

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$\frac{m_{1}}{m_{2}} = + 1$

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

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

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Solution

The correct option is D
$m_{1} . m_{2} = - 1$

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 $m_{1} = s l o p e o f ¯ ¯¯¯¯¯¯ ¯ R S a n d m_{2} = s l o p e o f ¯ ¯¯¯¯¯¯ ¯ O P$

In the figure, line l touches the circle with center O at point P. Q is the midpoint of radius OP. RS is a chord through Q such that chords RS || line l. If RS = 12 find the radius of the circle.