Question

A square has verticles O(0,0), A(a,o), B(a,a) and C(o,a).
Find:
(a) the mid point of the diagonals OB and CA
(b) the length of a diagonal of the square and the radius of the circle in which OABC is inscribed
(c) the equation of the circle inscribing the square.

Answer 1

Answer

(a) $D(\frac{a}{2};\frac{a}{2});$

(b) $OB=a\sqrt{2};$ $r=\frac{a}{\sqrt{2}};$

(c) $(x-\frac{a}{2})^2+(y-\frac{a}{2})^2=\frac{a^2}{2}$.

Explanation

(a) Because the vertices O(0,0), A(a,o), B(a,a) and C(o,a) are the vertices of the square, then o=0. The coordinates of the mid point the diagonals OB and CA are equal:

$x=\frac{(0+a)}2 = \frac{a}2$, $y=\frac{(0+a)}2 = \frac{a}2.$ Thus, the mid point of the diagonals is $D(\frac{a}2;\frac{a}2)$.

(b) The length of a diagonal of the square is equal to

$OB=CA=\sqrt{(a-0)^2+(a-0)^2}=a\sqrt{2}.$

The radius of the circle in which OABC is inscribed will be

$r=\frac{OB}2=\frac{a}{\sqrt{2}}.$

(c) The equation of circle with center C(x₀; y₀) and radius r is written as

$(x-x_0)^2+(y-y_0)^2=r^2.$

Because $D(\frac{a}2;\frac{a}2)$ is the center of the circle in which OABC is inscribed, and the $r=\frac{a}{\sqrt{2}}$ is its radius, then the equation of the circle inscribing the square is written as

$(x-\frac{a}2)^2+(y-\frac{a}2)^2=\frac{a^2}2.$