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(D) The equation of the radical axis of the two circles is obtained by subtracting the two equations: $(x^2+y^2+ax+by+c) - (x^2+y^2+bx+ay+c) = 0$.This

Vedclass · Accepted Answer

Statement-$I$ is false,Statement-$II$ is true.

Vedclass · Answer

(D) The equation of the radical axis of the two circles is obtained by subtracting the two equations: $(x^2+y^2+ax+by+c) - (x^2+y^2+bx+ay+c) = 0$.This simplifies to $(a-b)x + (b-a)y = 0$,which is $(a-b)(x-y) = 0$.Assuming $a 
eq b$,the radical axis is the line $x-y=0$,or $y=x$.Substituting $y=x$ into the first circle equation: $x^2+x^2+ax+bx+c=0 \Rightarrow 2x^2+(a+b)x+c=0$.Let the roots be $x_1$ and $x_2$. Then $x_1+x_2 = -\frac{a+b}{2}$ and $x_1x_2 = \frac{c}{2}$.The points of intersection are $(x_1, x_1)$ and $(x_2, x_2)$.The length of the common chord is the distance between these points: $\sqrt{(x_2-x_1)^2 + (x_2-x_1)^2} = \sqrt{2(x_2-x_1)^2} = \sqrt{2} |x_2-x_1|$.Using $|x_2-x_1| = \sqrt{(x_1+x_2)^2 - 4x_1x_2} = \sqrt{\frac{(a+b)^2}{4} - 2c} = \frac{\sqrt{(a+b)^2-8c}}{2}$.Thus,the length is $\sqrt{2} \cdot \frac{\sqrt{(a+b)^2-8c}}{2} = \frac{\sqrt{(a+b)^2-8c}}{\sqrt{2}}$.Therefore,Statement-$I$ is false.Statement-$II$ is a standard theorem in geometry of circles,which is true.Hence,Statement-$I$ is false and Statement-$II$ is true.

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