Question

Let any circle $S$ passes through the point of intersection of lines $\sqrt{3}(y-1)=x-1$ and $y-1=\sqrt{3}(x-1)$ and having its centre on the acute angle bisector of the given lines. If the common chord of $S$ and the circle $x^2+y^2+4x-6y+5=0$ passes through a fixed point, then the fixed point is

• A. $(\frac{1}{3},\frac{3}{2})$
• B. $(\frac{1}{2},\frac{3}{2})$
• C. $(\frac{1}{2},\frac{3}{4})$
• D. $(\frac{3}{2},\frac{3}{2})$

Answer 1

The correct option is B $(\frac{1}{2},\frac{3}{2})$

Given lines are
$\sqrt{3}(y-1)=x-1\Rightarrow x-\sqrt{3}y+\sqrt{3}-1=0\cdots \left(1\right)y-1=\sqrt{3}(x-1)\Rightarrow -\sqrt{3}x+y+\sqrt{3}-1=0\cdots \left(2\right)$

$a_1{a}_2+b_1{b}_2=-\sqrt{3}-\sqrt{3}<0$
So, the acute angle bisector is
$\frac{x-\sqrt{3}y+\sqrt{3}-1}{2}=+\frac{(-\sqrt{3}x+y+\sqrt{3}-1)}{2}\Rightarrow y=x$
Let the centre of the circle be $(\alpha ,\alpha )$

Point of intersection of $\left(1\right)$ and $\left(2\right)$, we get
$-2y+2=0\Rightarrow y=1\Rightarrow x=1$

Now, the equation of the circle is,
$(x-\alpha {)}^2+(y-\alpha {)}^2=(1-\alpha {)}^2+(1-\alpha {)}^2\Rightarrow x^2+y^2-2\alpha x-2\alpha y+4\alpha -2=0$
Given circle is
$x^2+y^2+4x-6y+5=0$

So, the common chord is,
$S_1-S_2=0\Rightarrow (4+2\alpha )x+(2\alpha -6)y+(7-4\alpha )=0\Rightarrow (4x-6y+7)+2\alpha (x+y-2)=0$
This represent family of lines, so the fixed point is the point of intersection of
$4x-6y+7=0\text{and}x+y-2=0\Rightarrow y=\frac{3}{2}\Rightarrow x=\frac{1}{2}$

Hence, the fixed point is
$(\frac{1}{2},\frac{3}{2})$