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In the given figure, PQ is a tangent to a circle with centre O$(-5, 3)$. If coordinates of P and Q are $(3, 1)$ and $(0, 6)$ respectively, then using distance formula, show that PQ $\perp$ OQ.
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$PQ^2 = 3^2 + (-5)^2 = 34$ (1/2 Mark)
$OP^2 = (-8)^2 + 2^2 = 68$ (1/2 Mark)
$OQ^2 = (-5)^2 + (-3)^2 = 34$ (1/2 Mark)
$OP^2=PQ^2+OQ^2 \Rightarrow PQ \perp OP$ (1/2 Mark)
$OP^2 = (-8)^2 + 2^2 = 68$ (1/2 Mark)
$OQ^2 = (-5)^2 + (-3)^2 = 34$ (1/2 Mark)
$OP^2=PQ^2+OQ^2 \Rightarrow PQ \perp OP$ (1/2 Mark)