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)?


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

$soit A (2, 1) B (3, 2) C (4, 3) points du$ $plan (ox, oy)$ $1) Determiner l^{'} equation du cercle qui$ $passe par A; B; C ?$ $2) points d intersection du cercle avec$ $l axe (ox, oy) ?$
	Answered by horsebrand11 last updated on 10/Jul/23
	$let(a,b) be the center point of the circle so { (((√((a−2)^2 +(b−1)^2 )) =(√((a−3)^2 +(b−2)^2 )))),(((√((a−2)^2 +(b−1)^2 ))=(√((a−4)^2 +(b−3)^2 )))) :} { ((−4a−2b+5=−6a−4b+13)),((−4a−2b+5=−8a−6b+25)) :} { ((2a+4b=8)),((4a+4b=20)) :}⇒ { ((a+2b=4)),((a+b=5)) :} { ((b=−1)),((a=6)) :} equation of the circle ⇒(x−6)^2 +(y+1)^2 =(6−2)^2 +(−1−1)^2 ⇒(x−6)^2 +(y+1)^2 =20$
	$let (a, b) be the center point of$ $the circle$ $so {\begin{cases} \sqrt{{(a - 2)}^{2} + {(b - 1)}^{2}} = \sqrt{{(a - 3)}^{2} + {(b - 2)}^{2}} \\ \sqrt{{(a - 2)}^{2} + {(b - 1)}^{2}} = \sqrt{{(a - 4)}^{2} + {(b - 3)}^{2}} \end{cases}$ ${\begin{cases} - 4 a - 2 b + 5 = - 6 a - 4 b + 13 \\ - 4 a - 2 b + 5 = - 8 a - 6 b + 25 \end{cases}$ ${\begin{cases} 2 a + 4 b = 8 \\ 4 a + 4 b = 20 \end{cases} \Rightarrow {\begin{cases} a + 2 b = 4 \\ a + b = 5 \end{cases}$ ${\begin{cases} b = - 1 \\ a = 6 \end{cases}$ $equation of the circle$ $\Rightarrow {(x - 6)}^{2} + {(y + 1)}^{2} = {(6 - 2)}^{2} + {(- 1 - 1)}^{2}$ $\Rightarrow {(x - 6)}^{2} + {(y + 1)}^{2} = 20$
	Commented by kapoorshah last updated on 10/Jul/23

	$w r o n g$ $B (3, 2) \Rightarrow {(3 - 6)}^{2} + {(2 + 1)}^{2} \neq 20$
	Answered by witcher3 last updated on 10/Jul/23

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