Given: Radius of the circle (OT or OQ) = 5 cmDistance PQ = xPT is a tangent to the circle at T.Since OT is the radius and PT is the tangent at T, we know that OT ⟂ PT. Therefore, ∆ OTP is a right-angled triangle at T.In ∆ OTP, using Pythagoras theorem:$OP^2=OT^2+PT^2$We can write OP as:$OP = OQ + PQ$$OP = 5 + x$Substituting the values in the Pythagoras equation:$(x+5)^2 = 5^2 + PT^2$$x^2 + 10 x+ 25 = 25 + PT^2$$PT^2 = x^2 + 10x + 25 - 25$$PT^2 = x^2 + 10x$