$A$ square is inscribed in the circle $x^2 + y^2 - 6x + 8y - 103 = 0$ with its sides parallel to the coordinate axes. Then the distance of the vertex of the square which is nearest to the origin is

  • A
    $6$
  • B
    $\sqrt{137}$
  • C
    $\sqrt{41}$
  • D
    $13$

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Similar Questions

Observe the following statements:
$I$. The circle $x^2+y^2-6x-4y-7=0$ touches the $y$-axis.
$II$. The circle $x^2+y^2+6x+4y-7=0$ touches the $x$-axis.
Which of the following is a correct statement?

Consider the circles ${x^2} + {(y - 1)^2} = 9$ and ${(x - 1)^2} + {y^2} = 25$. They are such that:

Let $C_1$ and $C_2$ be two circles touching each other externally at point $A$. Let $AB$ be the diameter of circle $C_1$. Draw a secant $BA_3$ to circle $C_2$,intersecting circle $C_1$ at point $A_1$ (where $A_1 \neq A$) and circle $C_2$ at points $A_2$ and $A_3$. If $BA_1 = 2$,$BA_2 = 3$,and $BA_3 = 4$,then the radii of circles $C_1$ and $C_2$ are respectively:

Consider a semicircle of radius $1$ unit constructed on the diameter $AB$ and let $O$ be its centre. Let $C$ be a point on $AO$ such that $AC:CO = 2:1$. Draw $CD$ perpendicular to $AO$ with $D$ on the semicircle. Draw $OE$ perpendicular to $AD$ with $E$ on $AD$. Let $OE$ and $CD$ intersect at $H$. Then,$DH$ equals

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