  Question

If the circle $$\mathrm{x}^{2}+\mathrm{y}^{2}+2\mathrm{g}\mathrm{x}+2\mathrm{f}\mathrm{y}+\mathrm{c}=0$$ bisects the circumference of the circle$$\mathrm{x}^{2}+\mathrm{y}^{2}+2\mathrm{g}_{1}\mathrm{x}+2\mathrm{f}_{1}\mathrm{y}+\mathrm{c}_{1}=0$$, then

A
2g1(gg1)+2f1(ff1)=cc1  B
2g1(gg1)+2f1(ff1)+cc1=0  C
g1(gg1)+f1(ff1)=cc1  D
2g(gg1)+2f1(ff1)=cc1  Solution

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)$$x^2+y^2 + 2g_1x + 2f_1y + c_1 = 0$$. (ii)The equation$$(x^2+ y^2 + 2gx + 2fy + c)-(x^2+y^2 + 2g_1x + 2f_1y + c_1)=0$$ (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$$x^2+y^2 + 2g_1x + 2f_1y + c_1 = 0$$ (ii)then $$(-g_1, —f_1)$$ lies on$$2g_1(g-g_1) +2f_1(f-f_1)=c-c_1$$Maths

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