Question

A circle $C$ of unit radius passing through the centers of the circles $C_1:x^2+y^2=1$ and $C_2:x^2+y^2=2x$ and having center on first quadrant. If $R,$ $R_1$ and $R_2$ are the radical axis of $C_1\&C_2;$ $C\&C_2;$ and $C\&C_1$ respectively, then which of the following is/are correct?

• A. $R_2:x+\sqrt{3}y=1$
• B. $\text{Radical Center}:(\frac{1}{2},\frac{1}{2\sqrt{3}})$
• C. $R,R_1\text{and}{R}_2\text{are concurrent}$
• D. $R_2:x-\sqrt{3}y=1$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $R_2:x+\sqrt{3}y=1$
B $\text{Radical Center}:(\frac{1}{2},\frac{1}{2\sqrt{3}})$
C $R,R_1\text{and}{R}_2\text{are concurrent}$

$C_1:x^2+y^2=1\Rightarrow \text{Center}:(0,0)C_2:x^2+y^2-2x=0\Rightarrow \text{Center}:(1,0)\text{Circle passes through two points}(0,0)\&(1,0)$
$(x-h{)}^2+(y-k{)}^2=1$
Putting $(0,0)\&(1,0)$,
$h=\frac{1}{2},k=\frac{\sqrt{3}}{2}$

$\text{Equation of circle is,}C:{(x-\frac{1}{2})}^2+{(y-\frac{\sqrt{3}}{2})}^2=1C:x^2+y^2-x-\sqrt{3}y=0$


The radical axis$\left(R_2\right)$ is given by
$R_2:C_1-C=0$
$x^2+y^2-1-(x^2+y^2-x-\sqrt{3}y)=0$
$\Rightarrow x+\sqrt{3}y=1$

Similarly, $R_1:C_2-C=0$
$\Rightarrow x-\sqrt{3}y=0$
Point of intersection of $R_1$ and $R_2$ is $(\frac{1}{2},\frac{1}{2\sqrt{3}})$

Similarly, $R:C_1-C_2=0$
$\Rightarrow x=\frac{1}{2}$

Therefore, $R,$ $R_1$ and $R_2$ are concurrent.