Question

The center of 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. (2/7, 18/7)
• B. (-2/7, 18/7)
• C. (1/7,- 9/7)
• D. (-1/7, 9/7)

Answer 1

Answer: (B)

(-2/7, 18/7)

Consider equation of first circle

$\therefore x^2+y^2+6x-1=0$

$\therefore x^2+y^2+2\left(3\right)x-1=0$

Comparing this equation with standard equation of circle i.e.$\therefore x^2+y^2+2gx+2fy+c=0$, we get,

$g_1=3$, $f_1=0$ and $c_1=-1$

Consider equation of second circle

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

$x^2+y^2+2\left(\frac{-3}{2}\right)y+2=0$

$\therefore g_2=0$, $f_2=\frac{-3}{2}$ and $c_2=2$

Consider equation of third circle.

$\therefore x^2+y^2+x+y-3=0$

$\therefore x^2+y^2+2\left(\frac{1}{2}\right)x+2\left(\frac{1}{2}\right)y-3=0$

$\therefore g_3=\frac{1}{2}$, $f_3=\frac{1}{2}$ and $c_3=-3$

Now, when one circle touches other circle orthogonally, we can write the equation as,

$g\acute{g}+f\acute{f}=c+\acute{c}$

For first circle, $g{g}_1+f{f}_1=c+c_1$

$\therefore 3g+\left(0\right)f=c+(-1)$

$\therefore 3g=c-1$

$\therefore c=3g+1$ Equation (1)

For second circle, $g{g}_2+f{f}_2=c+c_2$

$\therefore \left(0\right)g+\left(\frac{-3}{2}\right)f=c+2$

$\therefore \frac{-3}{2}f=c+2$

From equation (1),

$\frac{-3}{2}f=3g+1+2$

$\therefore \frac{-3}{2}f=3g+3$

$\therefore -3f=2(3g+3)$

$\therefore -3f=6g+6$

$\therefore 6g+3f=-6$

Dividing both sides by 3, we get,

$\therefore 2g+f=-2$ Equation (2)

For third circle, $g{g}_3+f{f}_3=c+c_3$

$\therefore \left(\frac{1}{2}\right)g+\left(\frac{1}{2}\right)f=c+(-3)$

$\therefore \left(\frac{g}{2}\right)+\left(\frac{f}{2}\right)=c-3$

$\therefore g+f=2(c-3)$

$\therefore g+f=2c-6$

From equation (1), we get,

$\therefore g+f=2(3g+1)-6$

$\therefore g+f=6g+2-6$

$\therefore g+f=6g-4$

$\therefore 5g-f=4$ Equation (3)

Adding equations (2) and (3), we get,

$7g=2$

$\therefore g=\frac{2}{7}$

Put this value in equation (3), we get,

$5\left(\frac{2}{7}\right)-f=4$

$\therefore f=\frac{10}{7}-4$

$\therefore f=\frac{-18}{7}$

Put these values in equation (1), we get,

$c=3\left(\frac{2}{7}\right)+1$

$c=\frac{6}{7}+1$

$\therefore c=\frac{13}{7}$

Thus, center of circle is given by,

$C(-g,-f)$

$\therefore C(-\frac{2}{7},\frac{18}{7})$