Mathematics · 2022
JEE Main · 28 June 2022, Shift 2 · Q23
If one of the diameters of the circle x^2+y^2-2 √ 2 x-6 √ 2 y+14=0 is a chord of the circle (x-2 √ 2 )^2+(y-2 √ 2 )^2= r^2, then the value of r^2 is…
If one of the diameters of the circle $\displaystyle x^2+y^2-2 \sqrt{2} x-6 \sqrt{2} y+14=0$ is a chord of the circle $\displaystyle (x-2 \sqrt{2})^2+(y-2 \sqrt{2})^2=\mathrm{r}^2$, then the value of $\displaystyle \mathrm{r}^2$ is equal to \_\_\_\_.
Official answer
10
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.