$A B C D$ is a square with side 16 units and $A$ is the origin. If the equation of the circle circumscribing the square $A B C D$ is $x^2+y^2=4 k(x+y)$, then $k=$

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Aicharya · Accepted Answer

$A B C D$ is a square with side 16 units let $a=16$ units Centre of the circle $O\left(\frac{a}{2}, \frac{a}{2}\right)$

$ A C=\sqrt{(a-0)^2+(a-0)^2}=\sqrt{2} a $ So, radius of circle $=\frac{A C}{2}=\frac{\sqrt{2} a}{2}=\frac{a}{\sqrt{2}}$ $\therefore$ Equation of circle $ \begin{array}{rlrl} & \left(x-\frac{a}{2}\right)^2+\left(y-\frac{a}{2}\right)^2 & =\left(\frac{a}{\sqrt{2}}\right)^2 \\ \Rightarrow x^2+\frac{a^2}{4}-a x+y^2+\frac{a^2}{4}-a y & =\frac{a^2}{2} \\ \Rightarrow & x^2+y^2-a x-a y =0 \\ \Rightarrow & x^2+y^2 =a(x+y) \\ \text { here } & a =16 \\ \Rightarrow & x^2+y^2 =16(x+y) \\ \Rightarrow & x^2+y^2 =4.4(x+y) \\ \text { by comparing } & k =4 \end{array} $

$A B C D$ is a square with side 16 units and $A$ is the origin. If the equation of the circle circumscribing…

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