From-the-standard-equation-of-a-circle-using-the-origin-0-0-we-deduced-the-eqution-x-a-2-y-b-2-r-2-to-x-2-y-2-r-2-In-what-terms-do-we-use-this-formular-

Question Number 166641 by pete last updated on 23/Feb/22

$$\mathrm{From}\:\mathrm{the}\:\mathrm{standard}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{a}\:\mathrm{circle}, \\ $$$$\mathrm{using}\:\mathrm{the}\:\mathrm{origin}\:\left(\mathrm{0},\mathrm{0}\right),\:\mathrm{we}\:\mathrm{deduced}\:\mathrm{the}\:\mathrm{eqution} \\ $$$$\left(\mathrm{x}−\mathrm{a}\right)^{\mathrm{2}} +\left(\mathrm{y}−\mathrm{b}\right)^{\mathrm{2}} =\mathrm{r}^{\mathrm{2}} \:\mathrm{to}\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{r}^{\mathrm{2}} . \\ $$$$\mathrm{In}\:\mathrm{what}\:\mathrm{terms}\:\mathrm{do}\:\mathrm{we}\:\mathrm{use}\:\mathrm{this}\:\mathrm{formular}? \\ $$

Commented by MJS_new last updated on 24/Feb/22

well, we use this formula every time we have a circle with center (0∣0) and radius r... the question seems strange to me.

$$\mathrm{well},\:\mathrm{we}\:\mathrm{use}\:\mathrm{this}\:\mathrm{formula}\:\mathrm{every}\:\mathrm{time}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{a}\:\mathrm{circle}\:\mathrm{with}\:\mathrm{center}\:\left(\mathrm{0}\mid\mathrm{0}\right)\:\mathrm{and}\:\mathrm{radius}\:{r}… \\ $$$$\mathrm{the}\:\mathrm{question}\:\mathrm{seems}\:\mathrm{strange}\:\mathrm{to}\:\mathrm{me}. \\ $$

Answered by alephzero last updated on 24/Feb/22

(x−a)^2 +(y−b)^2 = r^2 ⇒ (y−b)^2 = r^2 −(x−a)^2 ⇒ y−b = ±(√(r^2 −(x−a)^2 )) ⇒ y = ±(√(r^2 −(x−a)^2 ))−b Plot of this equation looks like a circle.

$$\left({x}−{a}\right)^{\mathrm{2}} +\left({y}−{b}\right)^{\mathrm{2}} \:=\:{r}^{\mathrm{2}} \\ $$$$\Rightarrow\:\left({y}−{b}\right)^{\mathrm{2}} \:=\:{r}^{\mathrm{2}} −\left({x}−{a}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:{y}−{b}\:=\:\pm\sqrt{{r}^{\mathrm{2}} −\left({x}−{a}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\:{y}\:=\:\pm\sqrt{{r}^{\mathrm{2}} −\left({x}−{a}\right)^{\mathrm{2}} }−{b} \\ $$$$\mathrm{Plot}\:\mathrm{of}\:\mathrm{this}\:\mathrm{equation}\:\mathrm{looks}\:\mathrm{like}\:\mathrm{a} \\ $$$$\mathrm{circle}. \\ $$

Commented by mr W last updated on 24/Feb/22

for me (x−a)^2 +(y−b)^2 = r^2 looks more like a circle than y = ±(√(r^2 −(x−a)^2 ))−b.

$${for}\:{me}\:\left({x}−{a}\right)^{\mathrm{2}} +\left({y}−{b}\right)^{\mathrm{2}} \:=\:{r}^{\mathrm{2}} \:{looks} \\ $$$${more}\:{like}\:{a}\:{circle}\:{than} \\ $$$${y}\:=\:\pm\sqrt{{r}^{\mathrm{2}} −\left({x}−{a}\right)^{\mathrm{2}} }−{b}. \\ $$

Commented by pete last updated on 24/Feb/22

$$\mathrm{Thanks}\:\mathrm{Sir} \\ $$

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