A circle C whose radius is 1 unit touches the x-axis at point A. The center Q of C lies in the first quadrant. The tangent from the origin O to the circle touches it at T and a point P lies on it such that ∆OAP is a right-angled triangle at A and its perimeter is 8 units.The length of PQ isThe equation of circle C isThe equation of tangent OT is

Given QT=QA=1.Let PQ = x.Then PT=x2−1ΔTQP and △APO are similar triangles. Then, OT=OA=x+1x2−1or   1+x+2(x+1)x2−1+x2−1=8or   x=53AP=83,OP=163 Let   ∠AOP=2θ.Then, sin⁡2θ=12 From ΔOAQ, tan⁡θ=1OA or   OA=1tan ⁡θ  Also,      sin ⁡2θ=2tan,⁡θ1+tan2 ⁡θ=12 ∴ tan⁡ θ=2−3Hence, OA=2+3Hence, the equation of circle is {x−(2+3)}2+(y−1)2=1.The equation of tangent OT is x−0cos ⁡2θ=y−0sin ⁡2θ or   x−3y=0