Umformung unverständlich (Kreisgleichung)

Question

Aufgabe:

Ich verstehe eine Umformung in einer Rechnung nicht.

Wie komme ich von:

$$(x-1)^2+y^2\geq 4*(x^2+(y-1)^2)$$

auf:

$$(x+\frac{1}{3})^2+(y-\frac{4}{3})^2 \leq \frac{8}{9}$$

Danke für die Hilfe:)

Answer 1

\(\begin{aligned}\left(x-1\right)^{2}+y^{2} & \geq4\left(x^{2}+\left(y-1\right)^{2}\right) & & \text{ausmultiplizieren}\\x^{2}-2x+1+y^{2} & \geq4x^{2}+4y^{2}-8y+4 & & |-x^{2}+2x-y^{2}-4\\-3 & \geq3x^{2}+2x+3y^{2}-8y & & |:3\\-1 & \geq x^{2}+\frac{2}{3}x+y^{2}-\frac{8}{3}y & & |+\left(\frac{1}{3}\right)^{2}+\left(\frac{4}{3}\right)^{2}\\\frac{8}{9} & \geq x^{2}+\frac{2}{3}x+\left(\frac{1}{3}\right)^{2}+y^{2}-\frac{8}{3}y+\left(\frac{4}{3}\right)^{2} & & \text{binomische Formel}\\\frac{8}{9} & \geq\left(x+\frac{1}{3}\right)^{2}+\left(y-\frac{4}{3}\right)^{2} & &\end{aligned}\)

Answer 2

\( (x-1)^{2}+y^{2} \geq 4 \cdot\left[x^{2}+(y-1)^{2}\right] \)

\( x^{2}-2 x+1+y^{2} \geq 4 \cdot\left[x^{2}+y^{2}-2 y+1\right] \)

\( x^{2}-2 x+1+y^{2} \geq 4 \cdot x^{2}+4 y^{2}-8 y+4 \)

\( 4 \cdot x^{2}+4 y^{2}-8 y+4 \leq x^{2}-2 x+1+y^{2} \mid-x^{2} \)

\( 3 \cdot x^{2}+4 y^{2}-8 y+4 \leq x^{2}-2 x+1+y^{2} \mid-y^{2} \)

\( 3 \cdot x^{2}+3 y^{2}-8 y+4 \leq-2 x+1 \mid+2 x-4 \)

\( 3 \cdot x^{2}+2 x+3 y^{2}-8 y \leq-3 \mid: 3 \)

\( x^{2}+\frac{2}{3} x+y^{2}-\frac{8}{3} y \leq-1 \mid+ \) quadratische Ergänzungen \( \left(\frac{1}{3}\right)^{2}+\left(\frac{4}{3}\right)^{2} \)

\( \left(x+\frac{1}{3}\right)^{2}+\left(y-\frac{4}{3}\right)^{2} \leq-1+\left(\frac{1}{3}\right)^{2}+\left(\frac{4}{3}\right)^{2} \)

\( \left(x+\frac{1}{3}\right)^{2}+\left(y-\frac{4}{3}\right)^{2} \leq-1+\frac{1}{9}+\frac{16}{9}=\frac{17}{9}-\frac{9}{9}=\frac{8}{9} \)

\( \left(x+\frac{1}{3}\right)^{2}+\left(y-\frac{4}{3}\right)^{2} \leq \frac{8}{9} \)