Question

The equation of a line inclined at an angle $\frac{\pi }{4}$ to the $x-$axis, such that the two circles $x^2+y^2=4,$ $x^2+y^2-10x-14y+65=0$ intercept equal lengths on it, is

• A. $2x-2y-3=0$
• B. $2x-2y+3=0$
• C. $x-y+6=0$
• D. $x-y-6=0$

Answer 1

The correct option is A $2x-2y-3=0$

Equation of the circles $x^2+y^2=4$ and $x^2+y^2-10x-14y+65=0$ have centres, radius as $(0,0),2$ and $(5,7),3$ respectively.

Let equation of line be $y-x=c\dots \left(i\right)$
Perpendicular from $(0,0)$ on line $\left(i\right)$ is $\left|\frac{-c}{\sqrt{2}}\right|=\frac{c}{\sqrt{2}}$
In $\Delta AON,\sqrt{2^2-{\left(\frac{c}{\sqrt{2}}\right)}^2}=AN$
Similarly in $\Delta CPM,\sqrt{3^2-{\left(\frac{2-c}{\sqrt{2}}\right)}^2}=CM$
Given, $AB=CD$
$\Rightarrow AN=CM$
$\Rightarrow 4-\frac{c^2}{2}=9-\frac{(2-c{)}^2}{2}$
$\Rightarrow c=-\frac{3}{2}$
Therefore, equation of line :
$y=x-\frac{3}{2}$
$\Rightarrow 2x-2y-3=0$