Question

Equation of the circle passing through the point of intersection of the circles $x ^{2}$ $+y^{2}=6$ and $x^{2}+y^{2}-6 x+8=0$ and the point $(1,1)$ is

• A. $x^{2}+y^{2}-6 x+4=0$
• B. $x^{2}+y^{2}-3 y+1=0$
• C. $x^{2}+y^{2}-3 x+1=0$
• D. $x^{2}+y^{2}-3 x+7=0$

Answer 1

Answer: (C)

Equation of the circle passing through the intersection is
$S_{1}+ kS _{2}=0$.
$\Rightarrow x ^{2}+ y ^{2}-6+ k \left( x ^{2}+ y ^{2}-6 x +8\right)=0$ . . . . . . . (1)
Since the required circle passes through $(1,1)$
$1+1-6+k(1+1-6+8)=0 \Rightarrow k=1$.
$\therefore$ From (1) equation is $x^{2}+y^{2}-3 x+1=0$.