Question

A square is inscribed in the circle $x^2+y^2-4x-6y-5=0$ whose sides are parallel to co-ordinate axes then vertices of square are

• A. $(5,0),(5,6),(-1,0),(-1,6)$
• B. $(5,1),(5,-6),(-1,1),(-1,6)$
• C. $(5,1),(5,6),(-1,0),(1,6)$
• D. $(0,5),(-6,5),(0,-1),(6,1)$

Answer 1

Answer: (A)

$(5,0),(5,6),(-1,0),(-1,6)$

$x^2+y^2-4x-6y-5=0$

$(x-2{)}^2-4+(y-3{)}^2-9-5=0$

$(x-2{)}^2+(y-3{)}^2=18$

centre $(2,+3)$ radius $=\sqrt{18}=3\sqrt{2}$

If a square is inscribe in circle,

$\Rightarrow$ diagonal of square =diameter

$\sqrt{2}a=2\left(radius\right)$

$2=2\times 3\sqrt{2}$

$\Rightarrow a=6$

$\therefore$ Side of the square is $6$

Since the side are parallel to $x$ and $y$

the vertices one equidistant from the centre.

$(2+3,+3+3),(2-3,+3+3),(2-3,+3-3)$ and $(2+3,+3-3)$

$\Rightarrow (5,6),(-1,6),(-1,0),(5,0)$