If angle between two tangents drawn from a point P to a circle of radius a and center O is 90°, then


Let us consider a circle with center O and tangents PT and PR and angle between them is 90° and radius of circle is a

To show :

Proof :

In OTP and ORP

TO = OR [ radii of same circle]

OP = OP [ common ]

TP = PR [ tangents through an external point to a circle are equal]

OTP ORP [ By Side Side Side Criterion ]

TPO = OPR [Corresponding parts of congruent triangles are equal ] [1]

Now, TPR = 90° [Given]

TPO + OPR = 90°

TPO + TPO = 90° [By 1]

TP0 = 45°

Now, OT TP [ As tangent at any point on the circle is perpendicular to the radius through point of contact]

OTP = 90°

So POT is a right-angled triangle

And we know that,


[As OT is radius and equal to a]

Hence Proved !