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A circle is inscribed (i.e. touches all four sides ) into a rhombous $ABCD$ with one angle $60^{0}$ . The distance from the centre of the circle to the nearest vertex is equal to $1$ . If P is any point of the circle, then $\left | PA \right |^{2}+\left | PB \right |^{2}+\left | PC \right |^{2}+\left | PD \right |^{2}$ is equal to:
Option: 1
0.12

Option: 2
11

Option: 3
10

Option: 4
13

Answers (1)

best_answer

Distance formula -

The distance between the point $A\left ( x_{1},y_{1} \right )\: and \: B\left ( x_{2},y_{2} \right )$

is $\sqrt{\left ( x_{1} -x_{2}\right )^{2}+\left ( y_{1} -y_{2}\right )^{2}}$

- wherein

Parametric form -

$x= r\cos \Theta$

$y= r\sin \Theta$

- wherein

Circle with centre $\left ( O,O \right )$ and radius $r$ .

$\because \:\:\tan 60^{0}=\frac{OA}{1}=\sqrt{3}$

$\because \:\:A(\sqrt{3,0})\:\:\:and\:\:\:C(-\sqrt{3},0)$

$\because \:\:\sin60^{0}=\frac{r}{1}=\frac{\sqrt{3}}{2}$

Let coordinates of any point $P$ on the circle be $P$ $\equiv (r\cos\theta,r\sin\theta)$

$\therefore PA^{2}=(\sqrt{3}-r \cos \theta)^{2}+(r \sin \theta)^{2}$

$PB^{2}=(r \cos \theta)^{2}+(1-r \sin \theta)^{2}$

$PC^{2}=(r \cos \theta+\sqrt{3})^{2}+(r \sin \theta)^{2}$

and $PD^{2}=(r \cos \theta)^{2}+(r \sin \theta+1)^{2}$

$\therefore PA^{2}+PB^{2}+PC^{2}+PD^{2}=4r^{2}+8=11$

$\because r=\sqrt{3}/2$

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manish painkra

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