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, $\frac{x+1}{h}=hx-h$
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 $h^3-3h^2+4=0\Rightarrow (h-2{)}^2(h+1)=0]$
$\therefore h=2$
Coordinates of the centre $Q$ are $(h,1)$ or $(2,1)$.
$\therefore$ The equation of the circle is $(x-2{)}^2+(y-$ $1{)}^2=1]$