Square inscribed in a circle (radius 4 cm) with an external tangent: PQRS is a square inscribed in a circle of radius 4 cm (so the circle's diameter equals the square's diagonal). Side PQ is extended beyond Q to meet a point Y. From Y, a tangent is.

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Solve this multiple-choice question and choose the correct option.

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Introduction / Context:This geometry problem mixes an inscribed square and a tangent from an external point. Converting the figure to coordinates makes the tangent-intersection computation straightforward and avoids guesswork. Given Data / Assumptions: Circle center O with radius R = 4 cm. PQRS is a square inscribed in the circle (its diagonal equals the circle’s diameter). PQ is extended to a point Y; from Y, the tangent to the circle touches at R. Required: length SY. Concept / Approach:Place the circle at the origin. Choose the square aligned with axes, so vertices are at (±2√2, ±2√2). The tangent at a point of the circle is perpendicular to the radius to that point. Intersect the tangent at R with the horizontal line containing PQ to get Y, then compute SY by distance formula. Step-by-Step Solution: Let P = (2√2, 2√2), Q = (−2√2, 2√2), R = (−2√2, −2√2), S = (2√2, −2√2).Line PQ: y = 2√2. Radius OR has slope 1 ⇒ tangent at R has slope −1 with equation y = −x − 4√2.Intersection with y = 2√2 gives 2√2 = −x − 4√2 ⇒ x = −6√2, so Y = (−6√2, 2√2).SY distance: Δx = −8√2, Δy = 4√2 ⇒ SY = √( (8√2)^2 + (4√2)^2 ) = √(128 + 32) = √160 = 4√10. Verification / Alternative check:Any square rotation about O keeps distances congruent. With symmetry, the computed value is invariant under rigid motions, confirming 4√10. Why Other Options Are Wrong:2√10 and 3√5 underestimate; 6√10 overestimates; 5√6 does not match the derived √160. Common Pitfalls:Using side as 4 (it is 4√2) or taking the tangent slope equal (not perpendicular) to the radius. Ensure tangent slope = −1 when radius slope = 1. Final Answer:4√10

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