Question

If the circle $x^2+y^2+2gx+2fy+c=0$ bisects the circumference of the circle $x^2+y^2+2g^{\prime }x+2f^{\prime }y+c^{\prime }=0$ then

• A. $2g^{\prime }(g-g^{\prime })+2f^{\prime }(f-f^{\prime })=c-c^{\prime }$
• B. $g^{\prime }(g-g^{\prime })+f^{\prime }(f-f^{\prime })=c-c^{\prime }$
• C. $2g(g-g^{\prime })+2f(f-f^{\prime })=c-c^{\prime }$
• D. None of these

Answer 1

Answer: (A)

$2g^{\prime }(g-g^{\prime })+2f^{\prime }(f-f^{\prime })=c-c^{\prime }$

The given circles are

$S_1:x^2+y^2+2gx+2fy+c=0$ ...(1)

and, $S_2:x^2+y^2+2g^{\prime }x+2f^{\prime }y+c^{\prime }=0$ ...(2)

The equation of common chord of (1) and (2) is

$S_1-S_2=0$

i.e., $2(g-g^{\prime })x+2(f-f^{\prime })y+(c-c^{\prime })=0$ ...(3)

Since (1) bisects the circumference of (2), therefore common chord will be the diameter of circle (2)

$\therefore$ Center $(-g^{\prime },-f^{\prime })$ of circle (2) lies on (3).

$\therefore -2(g-g^{\prime })g^{\prime }-2(f-f^{\prime })f^{\prime }+c-c^{\prime }=0$

or, $2g^{\prime }(g-g^{\prime })+2f^{\prime }(f-f^{\prime })=c-c^{\prime }.$