Let the lines y+2x=sqrt{11}+7sqrt{7} and 2y+x | vedclass

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(B) The normals to the circle are given by the equations:$y+2x=\sqrt{11}+7\sqrt{7}$  $(i)$$2y+x=2\sqrt{11}+6\sqrt{7}$  $(ii)$Solving these equations f

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$816$

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(B) The normals to the circle are given by the equations:$y+2x=\sqrt{11}+7\sqrt{7}$  $(i)$$2y+x=2\sqrt{11}+6\sqrt{7}$  $(ii)$Solving these equations for the intersection point $(h, k)$,which is the center of the circle:From $(i)$,$y = \sqrt{11}+7\sqrt{7}-2x$.Substituting into $(ii)$: $2(\sqrt{11}+7\sqrt{7}-2x)+x = 2\sqrt{11}+6\sqrt{7}$$2\sqrt{11}+14\sqrt{7}-4x+x = 2\sqrt{11}+6\sqrt{7}$$-3x = -8\sqrt{7} \implies x = h = \frac{8\sqrt{7}}{3}$Substituting $h$ back into $(i)$: $y = k = \sqrt{11}+7\sqrt{7}-2(\frac{8\sqrt{7}}{3}) = \sqrt{11} + \frac{21\sqrt{7}-16\sqrt{7}}{3} = \sqrt{11} + \frac{5\sqrt{7}}{3}$.Thus,the center is $(h, k) = (\frac{8\sqrt{7}}{3}, \sqrt{11}+\frac{5\sqrt{7}}{3})$.The radius $r$ is the perpendicular distance from the center $(h, k)$ to the tangent line $\sqrt{11}y-3x-(\frac{5\sqrt{77}}{3}+11) = 0$.$r = \frac{|\sqrt{11}(\sqrt{11}+\frac{5\sqrt{7}}{3}) - 3(\frac{8\sqrt{7}}{3}) - (\frac{5\sqrt{77}}{3}+11)|}{\sqrt{(\sqrt{11})^{2}+(-3)^{2}}}$$r = \frac{|11 + \frac{5\sqrt{77}}{3} - 8\sqrt{7} - \frac{5\sqrt{77}}{3} - 11|}{\sqrt{11+9}} = \frac{|-8\sqrt{7}|}{\sqrt{20}} = \frac{8\sqrt{7}}{2\sqrt{5}} = 4\sqrt{\frac{7}{5}}$.Then $r^{2} = 16 	imes \frac{7}{5} = \frac{112}{5}$.Now,calculate $(5h-8k)^{2}+5r^{2}$:$5h = 5(\frac{8\sqrt{7}}{3}) = \frac{40\sqrt{7}}{3}$$8k = 8(\sqrt{11}+\frac{5\sqrt{7}}{3}) = 8\sqrt{11} + \frac{40\sqrt{7}}{3}$$5h-8k = -8\sqrt{11} \implies (5h-8k)^{2} = 64 	imes 11 = 704$.$5r^{2} = 5 	imes \frac{112}{5} = 112$.$(5h-8k)^{2}+5r^{2} = 704 + 112 = 816$.

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

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