The equation of the circle whose diameter is the common chord of the circles $x^2+y^2+2 x+3 y+2=0$ and…

The equation of the circle whose diameter is the common chord of the circles $x^2+y^2+2 x+3 y+2=0$ and $x^2+y^2+2 x-3 y-4=0$ is
  1. $x^2+y^2+2 x+2 y+2=0$
  2. $x^2+y^2+2 x+2 y-1=0$
  3. $x^2+y^2+2 x+2 y+1=0$
  4. $x^2+y^2+2 x+2 y+3=0$

Solution

Given that
Equation of common chord is $\begin{aligned} S_1-S_2 & =0 \\ \Rightarrow \quad 6 y+6=0 \Rightarrow y & =-1 \end{aligned}$ Putting $y=-1$ in Eq. (i), we get $\begin{array}{rlrl} & \therefore & & x^2+1+2 x-3+2=0 \\ \Rightarrow & & x^2+2 x=0 \Rightarrow x=0,-2 \end{array}$ $\therefore$ End points of diameter are $(0,-1) \text { and }(-2,-1)$ Equation of circle is $\begin{aligned} (x-0) & (x+2)+(y+1)(y+1) & =0 \\ \Rightarrow & x^2+2 x+y^2+2 y+1 & =0 \end{aligned}$

Asked in: AP EAMCET 2005

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