The radical axis of circles $x^2+y^2+5 x+4 y-5=0$ and $x^2+y^2-3 x+5 y-6=0$ is

The radical axis of circles $x^2+y^2+5 x+4 y-5=0$ and $x^2+y^2-3 x+5 y-6=0$ is
  1. $8 y-x+1=0$
  2. $8 x-y+1=0$
  3. $8 x-8 y+1=0$
  4. $y-8 x+1=0$

Solution

Let $S_1 \equiv x^2+y^2+5 x+4 y-5=0$ and $S_2=x^2+y^3-3 x+5 y-6=0$ The radical axis is $\quad S_1-S_3=0$ $\begin{array}{rrr}\Rightarrow & \left(x^2+y^2+5 x+4 y-5\right) \\ & -\left(x^2+y^2-3 x+5 y-6\right)=0 \\ \Rightarrow & 5 x+3 x+4 y-5 y-5+6=0 \\ \Rightarrow & 8 x-y+1=0\end{array}$

Asked in: AP EAMCET 2001

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