soit-A-2-1-B-3-2-C-4-3-points-du-plan-ox-oy-1-Determiner-l-equation-du-cercle-qui-passe-par-A-B-C-2-points-d-intersection-du-cercle-avec-l-axe-ox-oy-

Question Number 194564 by a.lgnaoui last updated on 10/Jul/23

$$\mathrm{soit}\:\:\:\boldsymbol{\mathrm{A}}\left(\mathrm{2},\mathrm{1}\right)\:\:\:\:\:\boldsymbol{\mathrm{B}}\left(\mathrm{3},\mathrm{2}\right)\:\:\:\:\:\boldsymbol{\mathrm{C}}\left(\mathrm{4},\mathrm{3}\right)\:\:\mathrm{points}\:\mathrm{du}\: \\ $$$$\:\:\mathrm{plan}\left(\:\mathrm{ox},\mathrm{oy}\right)\:\: \\ $$$$ \\ $$$$\left.\:\mathrm{1}\right)\mathrm{Determiner}\:\:\mathrm{l}\:'\:\mathrm{equation}\:\mathrm{du}\:\mathrm{cercle}\:\mathrm{qui}\: \\ $$$$\:\:\:\:\mathrm{passe}\:\mathrm{par}\:\boldsymbol{\mathrm{A}};\:\boldsymbol{\mathrm{B}};\:\boldsymbol{\mathrm{C}}\:\:\:? \\ $$$$\left.\:\:\mathrm{2}\right)\:\mathrm{points}\:\mathrm{d}\:\mathrm{intersection}\:\mathrm{du}\:\mathrm{cercle}\:\mathrm{avec} \\ $$$$\:\:\:\:\:\mathrm{l}\:\mathrm{axe}\left(\mathrm{ox},\mathrm{oy}\right)? \\ $$

Answered by horsebrand11 last updated on 10/Jul/23

let(a,b) be the center point of the circle so { (((√((a−2)^2 +(b−1)^2 )) =(√((a−3)^2 +(b−2)^2 )))),(((√((a−2)^2 +(b−1)^2 ))=(√((a−4)^2 +(b−3)^2 )))) :} { ((−4a−2b+5=−6a−4b+13)),((−4a−2b+5=−8a−6b+25)) :} { ((2a+4b=8)),((4a+4b=20)) :}⇒ { ((a+2b=4)),((a+b=5)) :} { ((b=−1)),((a=6)) :} equation of the circle ⇒(x−6)^2 +(y+1)^2 =(6−2)^2 +(−1−1)^2 ⇒(x−6)^2 +(y+1)^2 =20

$$\mathrm{let}\left(\mathrm{a},\mathrm{b}\right)\:\mathrm{be}\:\mathrm{the}\:\mathrm{center}\:\mathrm{point}\:\mathrm{of} \\ $$$$\:\mathrm{the}\:\mathrm{circle}\: \\ $$$$\:\mathrm{so}\:\begin{cases}{\sqrt{\left(\mathrm{a}−\mathrm{2}\right)^{\mathrm{2}} +\left(\mathrm{b}−\mathrm{1}\right)^{\mathrm{2}} }\:=\sqrt{\left(\mathrm{a}−\mathrm{3}\right)^{\mathrm{2}} +\left(\mathrm{b}−\mathrm{2}\right)^{\mathrm{2}} }}\\{\sqrt{\left(\mathrm{a}−\mathrm{2}\right)^{\mathrm{2}} +\left(\mathrm{b}−\mathrm{1}\right)^{\mathrm{2}} }=\sqrt{\left(\mathrm{a}−\mathrm{4}\right)^{\mathrm{2}} +\left(\mathrm{b}−\mathrm{3}\right)^{\mathrm{2}} }}\end{cases} \\ $$$$\:\:\begin{cases}{−\mathrm{4a}−\mathrm{2b}+\mathrm{5}=−\mathrm{6a}−\mathrm{4b}+\mathrm{13}}\\{−\mathrm{4a}−\mathrm{2b}+\mathrm{5}=−\mathrm{8a}−\mathrm{6b}+\mathrm{25}}\end{cases} \\ $$$$\:\:\begin{cases}{\mathrm{2a}+\mathrm{4b}=\mathrm{8}}\\{\mathrm{4a}+\mathrm{4b}=\mathrm{20}}\end{cases}\Rightarrow\begin{cases}{\mathrm{a}+\mathrm{2b}=\mathrm{4}}\\{\mathrm{a}+\mathrm{b}=\mathrm{5}}\end{cases} \\ $$$$\:\:\begin{cases}{\mathrm{b}=−\mathrm{1}}\\{\mathrm{a}=\mathrm{6}}\end{cases} \\ $$$$\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\: \\ $$$$\:\:\Rightarrow\left(\mathrm{x}−\mathrm{6}\right)^{\mathrm{2}} +\left(\mathrm{y}+\mathrm{1}\right)^{\mathrm{2}} =\left(\mathrm{6}−\mathrm{2}\right)^{\mathrm{2}} +\left(−\mathrm{1}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\:\:\Rightarrow\left(\mathrm{x}−\mathrm{6}\right)^{\mathrm{2}} +\left(\mathrm{y}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{20} \\ $$

Commented by kapoorshah last updated on 10/Jul/23

$${wrong} \\ $$$${B}\:\left(\mathrm{3},\:\mathrm{2}\right)\:\Rightarrow\:\left(\mathrm{3}−\mathrm{6}\right)^{\mathrm{2}} \:+\:\left(\mathrm{2}+\mathrm{1}\right)^{\mathrm{2}} \:\neq\:\mathrm{20} \\ $$

Answered by witcher3 last updated on 10/Jul/23

C∈[AB] the middle of [AB] no cercl existe in this Way intersection of line and circle can bee 0,1 or 2 points no 3

$$\mathrm{C}\in\left[\mathrm{AB}\right]\:\:\:\mathrm{the}\:\mathrm{middle}\:\mathrm{of}\:\left[\mathrm{AB}\right] \\ $$$$\mathrm{no}\:\mathrm{cercl}\:\mathrm{existe}\:\mathrm{in}\:\mathrm{this}\:\mathrm{Way} \\ $$$$\mathrm{intersection}\:\mathrm{of}\:\mathrm{line}\:\mathrm{and}\:\mathrm{circle}\:\mathrm{can}\:\mathrm{bee}\:\mathrm{0},\mathrm{1}\:\mathrm{or}\:\mathrm{2}\:\mathrm{points}\:\mathrm{no}\:\mathrm{3} \\ $$$$ \\ $$

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