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  • The centers of the circles <img alt="\mathrm{C_1: x^2+y^2=4} \; and\; {C_2: x^2+y^2-8 x+7=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BC_1%3A%20x%5E2&plus;y%5E2%3D4

The centers of the circles \mathrm{C_1: x^2+y^2=4} \; and\; {C_2: x^2+y^2-8 x+7=0}are taken as the end points of a diameter of circle C. Then, the points of intersection of the common chord of circles

\mathrm{C_{1}} and \mathrm{C_{2}} with circle C are

Option: 1

\mathrm{\left(\frac{11}{8}, \frac{\sqrt{131}}{8}\right)}


Option: 2

\mathrm{\left(\frac{11}{8}, \frac{\sqrt{231}}{8}\right)}


Option: 3

\left(\frac{11}{8}, \frac{-\sqrt{231}}{8}\right)


Option: 4

\left(\frac{11}{8}, \frac{-\sqrt{231}}{8}\right)


Answers (1)

best_answer

The centres of \mathrm{C_1}and \mathrm{C_2} are (0, 0) and (4, 0)

Equation of circle C is x \mathrm{(x-4)+y^2=0}

 \mathrm{\text { Or } C: x^2+y^2-4 x=0}      -----------(1)

The common chord of circles \mathrm{C_1}and \mathrm{C_2} is (Radical Axis)

\mathrm{ x^2+y^2-4=x^2+y^2-8 x+7\: \: Or\: \: 8 x=11}--(2)

On substituting (2) in (1), we obtain

\mathrm{y^2+\frac{121}{64}-\frac{44}{8}=0 \text { or } y^2=\frac{231}{64} \Rightarrow y= \pm \frac{\sqrt{231}}{8}}

Hence, the points are \mathrm{\left(\frac{11}{8}, \pm \frac{\sqrt{231}}{8}\right)}

Posted by

Ritika Kankaria

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