Question

A circle $S$, whose radius is $1$ unit, touches the $X$-axis at point $A$. The centre $Q$ of $S$ lies in the first quadrant. The tangent from the origin $O$ to the circle touches it at $A$ and its perimeter is $8$ unit.
Equation of the circle $S$ is :

• A. $(x-2)^{2}+(y-1)^{2}=1$
• B. $[ x -(2+\sqrt{3})]^{2}+( y -1)^{2}=1$
• C. $[x-(2-\sqrt{3})]^{2}+(y-1)^{2}=1$
• D. $(x-\sqrt{3})^{2}+(y-1)^{2}=1$

Answer 1

Answer: (A)

Let $OA = h$ and $PQ = x$.
Triangles OAP and QTP are similar
$\therefore \frac{ AP }{ PT }=\frac{ OA }{ QT }=\frac{ OP }{ PQ } $
$\Rightarrow \frac{ x +1}{ PT }=\frac{ h }{1}=\frac{ OP }{ x }$ ... (i)
From Equstion (i), we get $OP = hx$ Perimeter of $\Delta OAP =8$
$\Rightarrow OA + AP + OP =8 $
$\Rightarrow ( h +1)( x +1)=8$
From Equation (i) $PT =\frac{ x +1}{ h }$
Again from Equation (i),
$OP = hx \Rightarrow OT + PT = hx$
or $PT = hx - h$
Thus, $\frac{x+1}{h}=h x-h $
or $h^{2}(x-1)=x+1$
$\Rightarrow h^{2}\left[\frac{8}{h+1}-2\right]=\frac{8}{h+1}$
[From Eq. (ii)]
or $\left.h^{3}-3 h^{2}+4=0 \Rightarrow (h-2)^{2}(h+1)=0\right]$
$\therefore h =2$
Coordinates of the centre $Q$ are $( h , 1)$ or $(2,1)$.
$\therefore $ The equation of the circle is $( x -2)^{2}+( y -$ $\left.1)^{2}=1\right]$