Question

Consider two circles $S_1=0$ and $S_2=0$, each of radius $1\text{unit}$ touching internally the sides of $△OAB$ and $△ABC$ respectively. If $O\equiv (0,0),A\equiv (0,4)$ and $B,C$ are the points on positive $x-$axis such that $OB<OC$, then which of the following is/are correct?

• A. The cosine of the angle between the pair of tangent drawn from $A$ to the circle $S_1$ is $\frac{4}{5}$
• B. The circumcentre of $△OAB$ is $(\frac{3}{2},3)$
• C. The length of tangent from $A$ to the circle $S_2=0$ is $\frac{9}{2}\text{units}$
• D. If one of the diameter of the circle $S_2=0$ is a chord to the circle $S_3=0$ whose centre is $(\frac{3}{2},3)$, then the radius of circle $S_3=0$ is $3$.

Answer 1

Answer: (A), (C), (D)

The correct options are
A The cosine of the angle between the pair of tangent drawn from $A$ to the circle $S_1$ is $\frac{4}{5}$
C The length of tangent from $A$ to the circle $S_2=0$ is $\frac{9}{2}\text{units}$
D If one of the diameter of the circle $S_2=0$ is a chord to the circle $S_3=0$ whose centre is $(\frac{3}{2},3)$, then the radius of circle $S_3=0$ is $3$.

Radii of both the circle is $1\text{unit}$
So, the center of $S_1=0$ is $C_1=(1,1)$ and $S_2$ is $C_2=(\alpha ,1)$
$A=(0,4)$, let $B=(b,0)$


Equation of $AB$ is
$\frac{x}{b}+\frac{y}{4}=1$
This is a tangent to $S_1=0$, then
$\frac{|\frac{1}{b}+\frac{1}{4}-1|}{\sqrt{\frac{1}{b^2}+\frac{1}{16}}}=1\Rightarrow {(\frac{1}{b}-\frac{3}{4})}^2=\frac{1}{b^2}+\frac{1}{16}\Rightarrow -\frac{3}{2b}+\frac{9}{16}=\frac{1}{16}\Rightarrow b=3\therefore B=(3,0)\cos \theta =\frac{OA}{AB}=\frac{4}{5}$
Circumcentre of $△OAB$ is midpoint of $A$ and $B$
$=(\frac{3}{2},2)$
As $AB$ is tangent to circle $S_2=0$, so
$|\frac{\alpha }{3}+\frac{1}{4}-1|=\sqrt{\frac{1}{9}+\frac{1}{16}}\Rightarrow |4\alpha -9|=5\Rightarrow 4\alpha =9\pm 5\Rightarrow \alpha =1,\frac{7}{2}$
As $\alpha \ne 1$, so $\alpha =\frac{7}{2}$
Equation of circle $S_2=0$ is
${(x-\frac{7}{2})}^2+(y-1{)}^2=1$
Length of tangent from $A(0,4)$ to $S_2=0$
$=\sqrt{{(0-\frac{7}{2})}^2+(4-1{)}^2-1}=\frac{9}{2}$

We know that, $BQ=BN=\frac{1}{2}$, so
$N=(\frac{7}{2},0)$
$\therefore$ The centre of circle $S_2=0$ is
$C_2=(\frac{7}{2},1)$


Any diameter of the circle $S_2=0$ has midpoint at centre $C_2$
So, Perpendicular distance from centre of $S_3=0$ meets at $C_2$
Let the radius of $S_3=0$ be $r$
In $△C_3{C}_{2}S$ is
$r^2=1^2+{(\frac{7}{2}-\frac{3}{2})}^2+(3-1{)}^2\therefore r=3$