The equation of the circle inscribed in a squ | vedclass

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(A) The distance $d$ between the parallel lines $x+y-2=0$ and $x+y-6=0$ is given by $d = \frac{|-2 - (-6)|}{\sqrt{1^2+1^2}} = \frac{4}{\sqrt{2}} = 2\s

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$2x^2+2y^2-2x-14y+21=0$

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(A) The distance $d$ between the parallel lines $x+y-2=0$ and $x+y-6=0$ is given by $d = \frac{|-2 - (-6)|}{\sqrt{1^2+1^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$.The radius $r$ of the inscribed circle is half the distance between the parallel lines,so $r = \frac{d}{2} = \sqrt{2}$.The center of the circle is the intersection of the midlines of the square. The midlines are $x+y-4=0$ and $x-y+3=0$.Solving these equations: $(x+y=4)$ and $(x-y=-3)$. Adding them gives $2x=1$,so $x=\frac{1}{2}$. Substituting $x$ gives $y=4-\frac{1}{2}=\frac{7}{2}$.Thus,the center is $(\frac{1}{2}, \frac{7}{2})$.The equation of the circle is $(x-\frac{1}{2})^2 + (y-\frac{7}{2})^2 = (\sqrt{2})^2$.$x^2 - x + \frac{1}{4} + y^2 - 7y + \frac{49}{4} = 2$.$x^2 + y^2 - x - 7y + \frac{50}{4} = 2$.$x^2 + y^2 - x - 7y + 12.5 - 2 = 0$.$x^2 + y^2 - x - 7y + 10.5 = 0$.Multiplying by $2$,we get $2x^2 + 2y^2 - 2x - 14y + 21 = 0$.

The equation of the circle inscribed in a square formed by the lines $x+y-2=0$,$x+y-6=0$,$x-y+1=0$ and $x-y+5=0$ is

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