Circles

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Step 1: Draw diagram - Circle with center O, radius = 5 cm, tangent PT = 12 cm Step 2: Use tangent property - Tangent is perpendicular to radius at point of contact Step 3: Apply Pythagoras theorem in

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Step 1: Given - AB = AC = 8 cm (tangents from external point are equal), ∠BAC = 60° Step 2: In triangle ABC, AB = AC, so it's isosceles ∠ABC = ∠ACB = (180° - 60°)/2 = 60° Therefore, triangle ABC

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Step 1: Draw perpendicular from center O to chord PQ, meeting at M Step 2: Perpendicular from center bisects the chord PM = MQ = 16/2 = 8 cm Step 3: Given - OM = 6 cm (distance from center to chord

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Step 1: Let O be common center, AB be chord of outer circle touching inner circle at P Step 2: Since AB touches inner circle at P, OP ⊥ AB OP = radius of inner circle = 3 cm Step 3: OP bisects chor

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Use this formula when: - Finding tangent length from an external point - Given distance from external point to center and radius Example: Point P is 13 cm from center O of a circle with radius 5 cm T

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Given: Circle with center O, external point P, tangents PA and PB To prove: PA = PB Proof: Step 1: Join OP, OA, OB Step 2: ∠OAP = 90° (tangent ⊥ radius) ∠OBP = 90° (tangent ⊥ radius) Step 3: In ri

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Step 1: For a quadrilateral circumscribed about a circle (tangential quadrilateral): Sum of opposite sides are equal AB + CD = BC + AD Step 2: Substitute given values: 6 + 4 = 7 + AD 10 =

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Step 1: For chord AB - let M be foot of perpendicular from O OM ⊥ AB, so AM = MB = 12/2 = 6 cm Given: OM = 8 cm Step 2: Find radius using right triangle OMA: OA² = OM² + AM² OA² = 8² + 6²