Question

The center of a circle which cuts $x^2+y^2+6x-1=0,x^2+y^2-3y+2=0$ and $x^2+y^2+x+y-3=0$ orthogonally is

• A. $(\frac{1}{7},\frac{9}{7})$
• B. $(\frac{-1}{7},\frac{-9}{7})$
• C. $(\frac{1}{7},\frac{-9}{7})$
• D. $(\frac{-1}{7},\frac{9}{7})$

Answer 1

Answer: (D)

The given circles
Let $S_1\equiv x^2+y^2+6x-1=0,$
Centre $C_1=(-3,0),\text{ Radius }{R}_1=\sqrt{10}$
$S_2\equiv x^2+y^2-3y+2=0,$
$C_2=(0,3/2),R_2=1/2$
$S_3=x^2+y^2+x+y-3=0$
$C_3=(-1/2,-1/2),R_3=\sqrt{7/2}$
Let $S\equiv x^2+y^2+2gx+2fy+c=0$ be equation of circle which cut orthogonal all three circles.
Then, by condition of orthogonality
$-2g(-3)+2(-f)(0)=c-1$
$\Rightarrow 6g=c-1$...(i)
$-2g(0)+2f(-3/2)=c+2$
$\Rightarrow -3f=c+2$...(ii)
$-2g(-1/2)+2f(1/2)=c-3$
$\Rightarrow g+f=c-3$...(iii)
On solving Eqs. (i), (ii) and (iii), we get
$g=1/7,f=-9/7$
So, centre is $(-g,-f)=(-1/7,9/7)$