A circle S, whose radius is 1 unit, touches X-axis at the point A. The center Q of S 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 the triangle OAP is a right-angled triangle at A and its perimeter is 8 unit. Then which of the following statements is /are correct?

9. Let OA = h and PQ = x.
Triangles OAP and QTP are similar

$\frac{AP}{PT}=\frac{OA}{QT}=\frac{OP}{PQ}\Rightarrow \frac{x+1}{PT}=\frac{h}{1}=\frac{OP}{x}\mathrm{.....}\left(1\right)$

We get  OP = hx 
From eq. (1),  $PT=\frac{x+1}{h}$
Perimeter of   $\Delta OAP=8$
$\begin{array}{l}\Rightarrow OA+AP+OP=8\\ \Rightarrow (h+1)(x+1)=\mathrm{8...}....\left(2\right)\end{array}$

Again From the eq. (1),  $OP=hx\Rightarrow OT+PT=hx$
Or  $PT=hx-h \Rightarrow TP=h(x-1)$  
Thus, $\frac{x+1}{h}=hx-h$ or $h^2(x-1)=x+1\Rightarrow h^2[\frac{8}{h+1}-2]=\frac{8}{h+1}$ [ From eq.(2)]
Or  $h^3-3h^2+4=0\Rightarrow {(h-2)}^2(h+1)=0$

$\therefore h=2$

(1) From Eq. (2), we have 

$\begin{array}{l}(h+1)(x+1)=8\\ \therefore x=\frac{5}{3}\Rightarrow PQ=\frac{5}{3}\end{array}$

(2) Coordinates of the centre Q are $(h,1)$ or $(2,1)$
The equation of the circle is  ${(x-2)}^2+{(y-1)}^2=1$
(3) $AP=1+x=1+\frac{5}{3}=\frac{8}{3}$

Slope of the line OT is  $\frac{AP}{OA}=\frac{4}{3}$
Equation of the line OT is $y=\frac{4}{3}x$
Tangent to the circle parallel to OA is y = 2.
Solving with  $y=\frac{4}{3}x$, we get the coordinates of  $R(\frac{3}{2},2)$
Equation of the circle passing through O and R is $(x-0)(x-\frac{3}{2})+(y-0)(y-2)+\lambda (4x-3y)=\mathrm{0.......}\left(3\right)$
It passes through  $Q(2,1)$, So , we get  $\lambda =0$

$\therefore R.circleis2x^2+2y^2-3x-4y=0$