Is $A$ the answer??

1 vote

Consider a circle with centre at origin and radius $2\sqrt{2}$. A square is inscribed in the circle whose sides are parallel to the $X$ an $Y$ axes. The coordinates of one of the vertices of this square are

- $(2, -2)$
- $(2\sqrt{2},-2)$
- $(-2, 2\sqrt{2})$
- $(2\sqrt{2}, -2\sqrt{2})$

0 votes

Answer $A$

Let radius of the circle be $r$, then:$$ r = 2\sqrt2$$

Draw perpendiculars from this point to $x$ and $y$ axes respectively and let them be $a$ and $b$.

Now,

$$a^2 + b^2 = r^2 \implies 2a^2 = r^2 \;\because (a = b)$$

$$\implies 2a^2 = (2\sqrt2)^2 = 4 \times 2 \implies a = \pm2$$

So, the coordinates are $(2, -2)$

$\therefore A$ is the correct option.

Let radius of the circle be $r$, then:$$ r = 2\sqrt2$$

Draw perpendiculars from this point to $x$ and $y$ axes respectively and let them be $a$ and $b$.

Now,

$$a^2 + b^2 = r^2 \implies 2a^2 = r^2 \;\because (a = b)$$

$$\implies 2a^2 = (2\sqrt2)^2 = 4 \times 2 \implies a = \pm2$$

So, the coordinates are $(2, -2)$

$\therefore A$ is the correct option.