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  • The length of the common chord of the circles <img alt="\mathrm{(x-1)^2+(y+1)^2=c^2 \text { and }(x+1)^2+(y-1)^2=c^2 \text { is }}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmath

The length of the common chord of the circles \mathrm{(x-1)^2+(y+1)^2=c^2 \text { and }(x+1)^2+(y-1)^2=c^2 \text { is }}

Option: 1

\mathrm{\sqrt{c^2-2}}


Option: 2

\mathrm{\frac{1}{2} \sqrt{c^2-2}}


Option: 3

2 \sqrt{c^2-2}


Option: 4

\frac{1}{4} \sqrt{c^2-2}


Answers (1)

best_answer

For \mathrm{(x-1)^2+(y+1)^2=c^2 \text { and }(x+1)^2+(y-1)^2=c^2} 

the equation of the common chord (radical axis) is

\mathrm{\begin{aligned} & (x-1)^2+(y+1)^2=(x+1)^2+(y-1)^2 \\ & \text { Or } y=x \end{aligned}}-----------(1)

The centre and radius of \mathrm{(x-1)^2+(y+1)^2=c^2} are \mathrm{C_1(1,-1) \text { and } c}

Perpendicular distance of \mathrm{C_1} from common chord AB 

\mathrm{\begin{aligned} & C_1 D=\frac{-1-1}{ \pm \sqrt{1+1}}=\frac{2}{\sqrt{2}}=\sqrt{2} \\ & \therefore \ln \Delta A D C_1, \\ & A D^2=A C_1^2-C_1 D^2 \\ & \quad=c^2-(\sqrt{2})^2=c^2-2 \\ & A B=2(A D)=2 \sqrt{c^2-2} \end{aligned}}

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