Q4 of 31 Page 209

Two radii OA and OB of a circle with centre O are perpendicular to each other. If two tangents drawn at the point A and B intersect each other at the point T, let us prove that AB = OT and they bisect each other at a right-angle.

Formula used.


⇒ Perpendicular of tangent through point of contact pass through centre of circle


⇒ Sum of all angles of quadrilateral is 360°


⇒ Diagonals of square bisect each other at 90°


Solution.



Join AB and OT


In quadrilateral OATB


As OA and OB are perpendicular to each other


∠AOB = 90°


If tangents from point A and B are drawn


Then;


∠OAT = ∠OBT = 90°


Sum of all angles of quadrilateral is 360°


∠OAT + ∠OBT + ∠AOB + ∠ATB = 360°


90° + 90° + 90° + ∠ATB = 360°


∠ATB = 360° - 270°


= 90°


All angles of quadrilateral are 90°


Hence quadrilateral can be either square or rectangle


OA = OB ∵ (Both are radius of same circle)


If adjacent sides are equal


Then given quadrilateral is a square


In square diagonals are equal


Hence AB = OT


And diagonal bisect each other at 90°


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