Question

lf $(1,2)$ is one limiting point and $x^2+y^2$ -6x- $8y+25=0$ is one circle of a coaxial system then the equation of the radical axis of the coaxial system is

• A. $x-y-5=0$
• B. $x+y+5=0$
• C. $x+y-20=0$
• D. $x+y-5=0$

Answer 1

The correct option is D $x+y-5=0$

Equation of limiting circle is, $(x-1{)}^2+(y-2{)}^2=0$
$\Rightarrow x^2+y^2-2x-4y+5=0=S$ (say)
and let given circle be $S^{\prime }\equiv x^2+y^2-6x-8y+25=0$
Hence equation of radical axis is give by, $S-S^{\prime }=0$
$\Rightarrow 4x+4y-20=0\Rightarrow x+y-5=0$