Question

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

• A. $x^2+y^2+3x-5=0$
• B. $x^2+y^2-4x+2=0$
• C. $x^2+y^2+6x-4=0$
• D. $x^2+y^2-4y-2=0$

Answer 1

Answer: (A)

$x^2+y^2+3x-5=0$

Let $S_1\equiv x^2+y^2-6x-8=0$
and $S_2\equiv x^2+y^2-6=0$
Now the equation of the circle passing through the point $(1,1)$ and the point of intersection of $S_1$ and $S_2$
$S_1+\lambda S_2=0$
$\Rightarrow$ $(x^2+y^2-6x-8)+\lambda (x^2+y^2-6)=0$
$\Rightarrow$ $(1+\lambda )x^2+(1+\lambda )y^2-6x+(-8-6\lambda )=\mathrm{0........}\left(i\right)$
Since eq. $\left(i\right)$ passes through the point $(1,1)$
$\because$ Put $x=1,y=1$ in eq. $\left(ii\right)$ we get
$(1+\lambda )+(1+\lambda )-6+(-8-6\lambda )=0$
$\Rightarrow$ $-4\lambda -12=0$
$\Rightarrow$ $\lambda =-3$
On putting the value of '$\lambda$' in Eq. $\left(i\right)$ we get
$-2x^2-2y^2-6x+10=0$
$\Rightarrow$ $x^2+y^2+3x-5=0$