Centre of a Circle Lies on the Bisector of Angle between Two Tangents
In the given ...
Question
In the given figure, XY and X1Y1 are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X1Y1 at B then ∠AOB is ___.
A
100∘
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B
60∘
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C
80∘
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D
90∘
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Solution
The correct option is D90∘
In the figure, if we join point 'O' and 'C'. Then we can see that OC = DA = EB, OE = CB and OD = CA ∵ OC = OD = OE [radius of circle] ∴ ODAC and OEBC are squares
∵ By Theorem - The centre of a circle lies on the bisector of the angle between two tangents drawn from a point outside it.
∴ AO bisects ∠CAD and BO bisects ∠CBE. This means AO and BO are diagonals of square ODAC and OBEC respectively. So, ∠AOC = 45∘ and ∠BOC = 45∘.