9:{"metadata":[["$","title","0",{"children":"iimkbmsat0025 - The pair of tangents drawn from an external point to a circle are perpendicular to each other and the length of each tangent (from the point to the circle) is 5 cm. The radius of the circle is | PYQs + Solutions | AfterBoards"}],["$","meta","1",{"name":"description","content":"undefined Let P be the external point, O be the center of the circ<=, A and B be the points where tan>=nts touch the circ<=, and r be the radius. Given: PA = PB = 5 cm and <= APB = 90°

Since tan>=nts from an external point are equal in <=ngth and perpendicular to the radius at the point of contact: PA OA and PB OB Therefore: <= PAO = <= PBO = 90°

In quadrilateral PAOB: <= PAO + <= APB + <= PBO + <= AOB = 360° 90° + 90° + 90° + <= AOB = 360° <= AOB = 90°

Since all ang<=s in quadrilateral PAOB are 90° and PA = PB (tan>=nt <=ngths), OA = OB (radii), the quadrilateral PAOB is a square. In a square, all sides are equal: PA = OA = r Therefore: r = 5 cm

The radius of the circ<= is 5 cm."}],["$","link","2",{"rel":"manifest","href":"https://www.afterboards.in/manifest.json","crossOrigin":"$undefined"}],["$","meta","3",{"name":"keywords","content":"IIM-K (BMSAT),QA,Geometry,Circles"}],["$","meta","4",{"name":"robots","content":"index, follow"}],["$","meta","5",{"name":"category","content":"education"}],["$","meta","6",{"name":"google-site-verification","content":"google"}],["$","meta","7",{"property":"og:title","content":"iimkbmsat0025 - The pair of tangents drawn from an external point to a circle are perpendicular to each other and the length of each tangent (from the point to the circle) is 5 cm. The radius of the circle is | PYQs + Solutions"}],["$","meta","8",{"property":"og:description","content":"undefined Let P be the external point, O be the center of the circ<=, A and B be the points where tan>=nts touch the circ<=, and r be the radius. Given: PA = PB = 5 cm and <= APB = 90°

Since tan>=nts from an external point are equal in <=ngth and perpendicular to the radius at the point of contact: PA OA and PB OB Therefore: <= PAO = <= PBO = 90°

In quadrilateral PAOB: <= PAO + <= APB + <= PBO + <= AOB = 360° 90° + 90° + 90° + <= AOB = 360° <= AOB = 90°

Since all ang<=s in quadrilateral PAOB are 90° and PA = PB (tan>=nt <=ngths), OA = OB (radii), the quadrilateral PAOB is a square. In a square, all sides are equal: PA = OA = r Therefore: r = 5 cm

The radius of the circ<= is 5 cm."}],["$","meta","9",{"property":"og:url","content":"/iimk-bmsat-past-year-questions/2025/QA/35"}],["$","meta","10",{"property":"og:site_name","content":"AfterBoards | JIPMAT Preparation"}],["$","meta","11",{"property":"og:image","content":"https://www.afterboards.in/pyp/AfterBoards%20CUET%20Past%20Year%20Papers%20with%20Solutions.webp"}],["$","meta","12",{"property":"og:type","content":"website"}],["$","meta","13",{"name":"twitter:card","content":"summary_large_image"}],["$","meta","14",{"name":"twitter:site","content":"afterboards.in"}],["$","meta","15",{"name":"twitter:title","content":"iimkbmsat0025 - The pair of tangents drawn from an external point to a circle are perpendicular to each other and the length of each tangent (from the point to the circle) is 5 cm. The radius of the circle is | PYQs + Solutions"}],["$","meta","16",{"name":"twitter:description","content":"undefined Let P be the external point, O be the center of the circ<=, A and B be the points where tan>=nts touch the circ<=, and r be the radius. Given: PA = PB = 5 cm and <= APB = 90°

Since tan>=nts from an external point are equal in <=ngth and perpendicular to the radius at the point of contact: PA OA and PB OB Therefore: <= PAO = <= PBO = 90°

In quadrilateral PAOB: <= PAO + <= APB + <= PBO + <= AOB = 360° 90° + 90° + 90° + <= AOB = 360° <= AOB = 90°

Since all ang<=s in quadrilateral PAOB are 90° and PA = PB (tan>=nt <=ngths), OA = OB (radii), the quadrilateral PAOB is a square. In a square, all sides are equal: PA = OA = r Therefore: r = 5 cm

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