Question

Circles $x^2+y^2=1$ and $x^2+y^2-4x-6y+12=0$ cut off equal intercepts on a line through the point $(1,1)$. The slope of the line is

• A. $1$
• B. $-1$
• C. $\frac{3}{2}$
• D. $-\frac{3}{2}$

Answer 1

The correct option is C $\frac{3}{2}$

Let the equation of the line be $y=mx+c$. Since it passes through $(1,1)$
$\therefore 1=m+c$ i.e., $c=1-m$
So that the line becomes $y=mx+1-m$
i.e., $mx-y-m+1=0$ ...(1)
Length of perpendicular from the centre$(0,0)$ of the first circle to the line (1)
$=\frac{|0-0-m+1|}{\sqrt{m^2+1}}=\frac{|1-m|}{\sqrt{m^2+1}}$
So, the length of the intercept
$=2\sqrt{1^2-\frac{{(1-m)}^2}{m^2+1}}=2\sqrt{\frac{2m}{m^2+1}}$
Also, the length of $\perp$ from the centre $(2,3)$ of second circle to the line (1)
$=\frac{|2m-3-m+1|}{\sqrt{m^2+1}}=\frac{|m-2|}{\sqrt{m^2+1}}$
$\therefore$ the length of the intercept in this case
$=2\sqrt{1-\frac{{(m-2)}^2}{m^2+1}}=2\sqrt{\frac{4m-3}{m^2+1}}$
Given, $2\sqrt{\frac{2m}{m^2+1}}=2\sqrt{\frac{4m-3}{m^2+1}}$
$\Rightarrow \frac{2m}{m^2+1}=\frac{4m-3}{m^2+1}$

$\Rightarrow 2m=3$

$\Rightarrow m=\frac{3}{2}$