Discussion Forum

If a variable circle S=0 touches  the line y=x and passes through  the point (0,0)  , then the fixed point that lies on the common chord of the circles  $x^{2}+y^{2}+6x+8y-7=0$ and S=0 is 

Answer:

Explanation:

 Let the equation of circle S=0, passes through the point (0,0) is

    S=$x^{2}+y^{2}+2gx+2fy=0$  ..........(i)

 Since , circle S=0  touches the line y=x , so

        $\frac{|-g+f|}{\sqrt{2}}=\sqrt{g^{2}+f^{2}}$

   $\Rightarrow$         $g^{2}+f^{2}-2gf=2(g^{2}+f^{2})$

$\Rightarrow$        $g^{2}+f^{2}+2gf$=0

 $\Rightarrow$       g+f=0 

 So, the  equation of circle S=0, becomes

  $x^{2}+y^{2}+2gx-2gy=0$.....(ii)

 Now, the equation of common chord of circles (ii) and $x^{2}+y^{2}+6x+8y-7=0$ is 

                 $(2g-6)x-(2g+8)y+7=0$

$\Rightarrow$   $2g(x-y)-(6x+8y-7)=0$   ........(iii)

 eq.(iii)  represents the family of lines , passes through the intersection of lines

               x-y=0  and 6x+8y-7=0

 And the point of intersection is $\left(\frac{1}{2},\frac{1}{2}\right)$

 Hence, option (a) is correct