Since \(\angle OPS = 90^\circ \) and \( \angle OAS = 90^\circ \), hence \( ASPO \) is a cyclic quad.

Since \( \angle PAB = 90^\circ - \angle PBA = \angle PQO = 180^\circ - \angle OQP \), hence \( APQO \) is a cyclic quad.

Hence, \(ASPQO \) is a cyclic quad. Thus, since \( \angle SAO = 90^\circ \), hence \( \angle SQO = 90 ^ \circ \). And since \( \angle AOQ = 90^\circ \) hence \( \angle ASQ = 90^ \circ \). So we have \(ASQO \) is a rectangle.

With O as the origin, let's assume the circle to be: x²+ y² = R² and point P to be (a,b). Likewise, let Q be (q,0) & S be (p,R). Now it will be adequate to show that p=q for ASQO to be a rectangle. The tangent at P is:ax+by=R² and this meets the tangent at A with p as given by: ap +bR=R² or p =R(R-b)/a. By similarity, we further claim that: R/(R+b) = q/a or q=aR/(R+b). Now p=q if R(R-b)/a = aR/(R+b) or if (R--b)(R+b) = a² or if R² - b² = a² or if a² + b² = R² which is true since P:(a,b) lies on the circle; hence p = q. In other words, ASQO is rectangular.

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Since \(\angle OPS = 90^\circ \) and \( \angle OAS = 90^\circ \), hence \( ASPO \) is a cyclic quad.

Since \( \angle PAB = 90^\circ - \angle PBA = \angle PQO = 180^\circ - \angle OQP \), hence \( APQO \) is a cyclic quad.

Hence, \(ASPQO \) is a cyclic quad. Thus, since \( \angle SAO = 90^\circ \), hence \( \angle SQO = 90 ^ \circ \). And since \( \angle AOQ = 90^\circ \) hence \( \angle ASQ = 90^ \circ \). So we have \(ASQO \) is a rectangle.

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With O as the origin, let's assume the circle to be: x²+ y² = R² and point P to be (a,b). Likewise, let Q be (q,0) & S be (p,R). Now it will be adequate to show that p=q for ASQO to be a rectangle. The tangent at P is:ax+by=R² and this meets the tangent at A with p as given by: ap +bR=R² or p =R(R-b)/a. By similarity, we further claim that: R/(R+b) = q/a or q=aR/(R+b). Now p=q if R(R-b)/a = aR/(R+b) or if (R--b)(R+b) = a² or if R² - b² = a² or if a² + b² = R² which is true since P:(a,b) lies on the circle; hence p = q. In other words, ASQO is rectangular.

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