Question

If the circle $x^2+y^2+2gx+2fy+c=0$ bisects the circumference of the circle
$x^2+y^2+2g_{1}x+2f_{1}y+c_1=0$, then

• A. $2g_1(g-g_1)+2f_1(f-f_1)=c-c_1$
• B. $2g_1(g-g_1)+2f_1(f-f_1)+c-c_1=0$
• C. $g_1(g-g_1)+f_1(f-f_1)=c-c_1$
• D. $2g(g-g_1)+2f_1(f-f_1)=c-c_1$

Answer 1

The correct option is A $2g_1(g-g_1)+2f_1(f-f_1)=c-c_1$

Let the equations of the two circles be

$x^2+y^2+2gx+2fy+c=0$, (i)

$x^2+y^2+2g_{1}x+2f_{1}y+c_1=0$. (ii)

The equation

$(x^2+y^2+2gx+2fy+c)-(x^2+y^2+2g_{1}x+2f_{1}y+c_1)=0$ (iii)

is satisfied by the coordinates of any point whose coordinates satisfy both the equations (i) and (ii) ; hence it is a locus through the common points of the two circles,

But the equation is equivalent to

$2(g-g_1)x+2(f-f_1)y+c-c_1=0,$

which is a straight line and is therefore the common chord.

Note i. If one circle bisects the circumference of another, their commonchord is a diameter of the latter circle.

Thus, if $x^2+y^2+2gx+2fy+c=0$ (i)

bisects the circumference of

$x^2+y^2+2g_{1}x+2f_{1}y+c_1=0$ (ii)

then $(-g_1,—f_1)$ lies on

$2g_1(g-g_1)+2f_1(f-f_1)=c-c_1$