Question

The circle which passes through the points of intersection of the circles $x^2+y^2-6x+8=0,x^2+y^2=6$ and the point $(1,1)$ is

• A. $x^2+y^2+3x+1=0$
• B. $x^2-y^2-3x+1=0$
• C. $x^2+y^2-3x+1=0$
• D. $x^2-y^2+3x+1=0$

Answer 1

The correct option is C $x^2+y^2-3x+1=0$

The circle is $S+\lambda S^{\prime }=0$
$\Rightarrow x^2+y^2-6x+8+\lambda (x^2+y^2-6)=0$
It passes through $(1,1)$
$\therefore 1+1-6+8+\lambda (1+1-6)=0$
$\Rightarrow \lambda =1$
The circle is $x^2+y^2-6x+8+x^2+y^2-6=0$
or $2x^2+2y^2-6x+2=0$
or $x^2+y^2-3x+1=0$