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  • If one of the diameters of the circle <img alt="\mathrm{x^{2}+y^{2}-2 \sqrt{2} x-6 \sqrt{2} y+14=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E%7B2%7D&plus;y%5E%7B2%7

If one of the diameters of the circle \mathrm{x^{2}+y^{2}-2 \sqrt{2} x-6 \sqrt{2} y+14=0} is a chord of the circle \mathrm{(x-2 \sqrt{2})^{2}+(y-2 \sqrt{2})^{2}=\mathrm{r}^{2}}, then the value of \mathrm{r}^{2} is equal to___________.

Option: 1

10


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{C_{1}: x^{2}+y^{2}-2 \sqrt{2} x-6 \sqrt{2} y+14=0} \\

\mathrm{C_{2}: x^{2}+y^{2}-4 \sqrt{2} x-4 \sqrt{2} y+\left(16-r^{2}\right)=0}

Common Chord is \mathrm{C_{1}-C_{2}=0}

\mathrm{2 \sqrt{2} x-2 \sqrt{2} y-2+r^{2}=0}

It is a diameter of \mathrm{C_{1}}, so it passes through its centre (\sqrt{2}, 3 \sqrt{2})

\mathrm{ 2 \sqrt{2}(\sqrt{2})-2 \sqrt{2}(3 \sqrt{2})-2+r^{2}=0} \\

\mathrm{\Rightarrow r^{2}=2-4+12 }\\

\mathrm{\Rightarrow r^{2}=10 }

Hence answer is 10

Posted by

vinayak

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