Question

$P(\sqrt{2},\sqrt{2})$ is a point on the circle $x^2+y^2=4$ and $Q$ is another point on the circle such that arc $PQ=\frac{1}{4}$ (circumference). The coordinates of $Q$ are

• A. $(-\sqrt{2},-\sqrt{2})$
• B. $(\sqrt{2},-\sqrt{2})$
• C. $(-\sqrt{2},\sqrt{2})$
• D. $(2,\sqrt{2})$

Answer 1

Answer: (B), (C)

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

Given $P(\sqrt{2};\sqrt{2})$ is a point on circle $x^2+y^2=4$ and Q is another point such that PQ$=\frac{1}{4}\left(circumference\right).$

$\therefore$ $\angle POQ={90}^{\circ }$, O is a center of the circle $x^2+y^2=4$

So, slope of $OP\times$ slope of $OQ=-1$

$1\times \frac{k}{h}=-1$

$k=-h$ ---(1)
In $\Delta OPQ,$
$PQ=\sqrt{2}OP=\sqrt{2}\times 2=2\sqrt{2}$
$(h-\sqrt{2}{)}^2+(-h-\sqrt{2}{)}^2=(2\sqrt{2}{)}^2$
$2(h^2+2)=8$
$h^2=2$

$h=\sqrt{2}$ or $h=-\sqrt{2}$

$\therefore k=-\sqrt{2}$ or $k=\sqrt{2}$

So, $Q(\sqrt{2},-\sqrt{2})$ or $Q(-\sqrt{2},\sqrt{2})$. These are the two points available.