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Two circles touching internally at O. OXY, OAB straight lines, the latter passing through the centres. Prove that OX : OY = OA : OB.
Given : Two circles touching internally
at ‘O' . The line OXY touches the circlesat X and Y . The line OAB passes throughthe centres of the circles.R. T. P.: OX : OY = OA : OBConstruction: Join AX and BY.Proof : OA is the diametre of the innercircle.∠OXA = 900 ( Angle in semicircle )......................( i )OB is the diameter of the outer circle.
∠OYB = 900 ( Angle in the semi circle...................( ii )
From ( i ) and ( ii ); ∠OXA = ∠OYB =900
i.e., The corresponding angles are equal. AX // BYIn ΔOBY; AX //BYOX/OY = OA/OB
i.e., OX : OY = OA : OB
Calucations of gradients find f Graph some level curve f=const. f=9x^2 = 4y^2
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2
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-2+-9
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Provide me some Example of in-equations.
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