True

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 : OP = a√2
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,

Hence Proved !