3. Zadatak

Jednadžbu kružnice napišimo u obliku $(x-p)^2+(y-q)^2=r^2$. Apscise zajedničkih točaka hiperbole i kružnice su rješenja jednadžbe $(x-p)^2+\left(\dfrac1x-q\right)^2=r^2$, ili nakon sređivanja,

$$x^4-2px^3+(p^2+q^2-r^2)x^2-2qx+1=0.\qquad(*)$$

Vièteove formule za jednadžbu $(*)$ glase:

$$\begin{aligned}A&=x_1+x_2+x_3+x_4=2p,\\B&=x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4=p^2+q^2-r^2,\\C&=x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4=2q,\\D&=x_1x_2x_3x_4=1.\end{aligned}$$

(a) Prema Vièovim formulama je $x_1x_2x_3x_4=1$.

Zbog $y_1y_2y_3y_4=\dfrac1{x_1}\cdot\dfrac1{x_2}\cdot\dfrac1{x_3}\cdot\dfrac1{x_4}=1$, vrijedi i $y_1y_2y_3y_4=1$.

(b) Imamo:

$$\begin{aligned}S&=\sum_{k=1}^4|OP_k|^2=|OP_1|^2+|OP_2|^2+|OP_3|^2+|OP_4|^2\\&=x_1^2+y_1^2+x_2^2+y_2^2+x_3^2+y_3^2+x_4^2+y_4^2\\&=(x_1^2+x_2^2+x_3^2+x_4^2)+\left(\frac1{x_1^2}+\frac1{x_2^2}+\frac1{x_3^2}+\frac1{x_4^2}\right).\end{aligned}$$

Kako je

$$x_1^2+x_2^2+x_3^2+x_4^2=(x_1+x_2+x_3+x_4)^2-2(x_1x_2+\ldots+x_3x_4)=A^2-2B;$$

i

$$\begin{aligned}&\frac1{x_1^2}+\frac1{x_2^2}+\frac1{x_3^2}+\frac1{x_4^2}\\&=\frac{x_1^2x_2^2x_3^2+x_1^2x_2^2x_4^2+x_1^2x_3^2x_4^2+x_2^2x_3^2x_4^2}{(x_1x_2x_3x_4)^2}\\&=(x_1x_2x_3)^2+(x_1x_2x_4)^2+(x_1x_3x_4)^2+(x_2x_3x_4)^2\\&=C^2-2(x_1^2x_2^2x_3x_4+x_1^2x_2x_3^2x_4+x_1x_2^2x_3^2x_4\\&\qquad+x_1^2x_2x_3x_4^2+x_1x_2^2x_3x_4^2+x_1x_2x_3^2x_4^2)\\&=C^2-2x_1x_2x_3x_4(x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4)\\&=C^2-2\cdot1\cdot B,\end{aligned}$$

slijedi

$$\begin{aligned}S&=(A^2-2B)+(C^2-2B)\\&=(4p^2-2p^2-2q^2+2r^2)+(4q^2-2p^2-2q^2+2r^2)=4r^2.\end{aligned}$$