A square is inscribed in the circle x^2+y^2-1 | vedclass

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(D) The equation of the circle is $x^2+y^2-10x-6y+30=0$.Rewriting it as $(x-5)^2+(y-3)^2 = 25+9-30 = 4 = 2^2$.Thus,the center is $(5, 3)$ and the radi

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$152$

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(D) The equation of the circle is $x^2+y^2-10x-6y+30=0$.Rewriting it as $(x-5)^2+(y-3)^2 = 25+9-30 = 4 = 2^2$.Thus,the center is $(5, 3)$ and the radius $R = 2$.Let the sides of the square be parallel to $y=x+c$ and $x+y+d=0$.The distance from the center $(5, 3)$ to these lines must be equal to the distance from the center to the sides of the square,which is $R/\sqrt{2} = 2/\sqrt{2} = \sqrt{2}$.For $y-x-c=0$: $\left|\frac{3-5-c}{\sqrt{1^2+(-1)^2}}\right| = \sqrt{2} \implies |c+2| = 2 \implies c=0$ or $c=-4$.For $x+y+d=0$: $\left|\frac{5+3+d}{\sqrt{1^2+1^2}}\right| = \sqrt{2} \implies |d+8| = 2 \implies d=-6$ or $d=-10$.The equations of the sides are $y=x$,$y=x-4$,$x+y=6$,and $x+y=10$.Solving these pairwise gives the vertices: $(5, 5), (3, 3), (5, 1), (7, 3)$.Calculating $\sum(x_i^2+y_i^2) = (25+25) + (9+9) + (25+1) + (49+9) = 50 + 18 + 26 + 58 = 152$.

$A$ square is inscribed in the circle $x^2+y^2-10x-6y+30=0$. One side of this square is parallel to $y=x+3$. If $(x_i, y_i)$ are the vertices of the square,then $\sum(x_i^2+y_i^2)$ is equal to:

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