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

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


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