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  • Let the lines  <img alt="\mathrm{y+2 x=\sqrt{11}+7 \sqrt{7} \text { and } 2 y+x=2 \sqrt{11}+6 \sqrt{7}}" src="/latex-image/?%5Cmathrm%7By&plus;2%20x%3D%5Csqrt%7B11%7D&plus;7%20%5Csqrt%

Let the lines  \mathrm{y+2 x=\sqrt{11}+7 \sqrt{7} \text { and } 2 y+x=2 \sqrt{11}+6 \sqrt{7}} be normal to a circle \mathrm{C:(x-h)^{2}+(y-k)^{2}=r^{2} \text {. If the line } \sqrt{11} y-3 x=\frac{5 \sqrt{77}}{3}+11} is tangent to the circle C, then the value of \mathrm{\left ( 5h-8k \right )^{2}+5r^{2}} is equal to____________.

Option: 1

816


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

Point of intersection will be centre

\mathrm{2x+y=\sqrt{11}+7\sqrt{7}}

\Rightarrow \mathrm{4x+2y=2\sqrt{11}+14\sqrt{7}----1}

      \mathrm{x+2y=2\sqrt{11}+6\sqrt{7}----2}
        ____________________________________________

(1) \: -\: (2)\Rightarrow \mathrm{3x=8\sqrt{7}\Rightarrow x= \frac{8\sqrt{7}}{3}},

                \mathrm{y=\sqrt{11}+7\sqrt{7}-\frac{16\sqrt{7}}{3}=\sqrt{11}+\frac{5\sqrt{7}}{3}}

\mathrm{r}= distance of centre from tangent
            \mathrm{=\left | \frac{3\left ( \frac{8\sqrt{7}}{3} \right )-\sqrt{11}\left ( \sqrt{11}+\frac{5\sqrt{7}}{3} \right )+\frac{5\sqrt{77}}{3}+11}{\sqrt{9+11}} \right |}


            \mathrm{=\frac{8\sqrt{7}}{2\sqrt{5}}=4\sqrt{\frac{7}{5}}}       

        \\ \mathrm{h=\frac{8\sqrt{7}}{3},k=\sqrt{11}+\frac{5\sqrt{7}}{3},r=4\sqrt{\frac{7}{5}}}\\ \mathrm{\left ( 5h-8k \right )^{2}+5r^{2}}\\ \mathrm{=\left ( \frac{40\sqrt{7}}{3}-8\sqrt{11}-40\frac{\sqrt{7}}{3} \right )^{2}+5\times 16\times \frac{7}{5}}\\ \mathrm{=64\times 11+112=816}

Posted by

Ramraj Saini

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