Two circles with centres O and O^{prime} of r | vedclass

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(D) Given,two circles have radii $OP = 3\, cm$ and $PO^{\prime} = 4\, cm$.These two circles intersect at $P$ and $Q$.Since $OP$ and $O^{\prime}P$ are

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$4.8$

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(D) Given,two circles have radii $OP = 3\, cm$ and $PO^{\prime} = 4\, cm$.These two circles intersect at $P$ and $Q$.Since $OP$ and $O^{\prime}P$ are radii to the points of contact of the tangents at $P$,and the problem states $OP$ and $O^{\prime}P$ are perpendicular to each other at $P$,we have $\angle OPO^{\prime} = 90^{\circ}$.In right-angled $	riangle OPO^{\prime}$,by Pythagoras theorem:$(OO^{\prime})^2 = (OP)^2 + (PO^{\prime})^2 = 3^2 + 4^2 = 9 + 16 = 25$.So,$OO^{\prime} = 5\, cm$.Let $PN$ be the perpendicular from $P$ to $OO^{\prime}$,where $N$ is the intersection point on $OO^{\prime}$.Let $ON = x$,then $NO^{\prime} = 5 - x$.In right-angled $	riangle OPN$,$(PN)^2 = (OP)^2 - (ON)^2 = 3^2 - x^2 = 9 - x^2$ ... $(i)$In right-angled $	riangle PNO^{\prime}$,$(PN)^2 = (PO^{\prime})^2 - (NO^{\prime})^2 = 4^2 - (5 - x)^2 = 16 - (25 + x^2 - 10x) = 16 - 25 - x^2 + 10x = 10x - 9 - x^2$ ... (ii)Equating $(i)$ and (ii):$9 - x^2 = 10x - 9 - x^2$$18 = 10x \Rightarrow x = 1.8\, cm$.Substituting $x = 1.8$ in $(i)$:$(PN)^2 = 9 - (1.8)^2 = 9 - 3.24 = 5.76$.$PN = \sqrt{5.76} = 2.4\, cm$.The common chord $PQ = 2 	imes PN = 2 	imes 2.4 = 4.8\, cm$.

Two circles with centres $O$ and $O^{\prime}$ of radii $3\, cm$ and $4\, cm$,respectively,intersect at two points $P$ and $Q$ such that $OP$ and $O^{\prime}P$ are tangents to the two circles. Find the length of the common chord $PQ$ (in $cm$).

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