Math Question and Answers


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 192204 by naka3546 last updated on 11/May/23

	Answered by a.lgnaoui last updated on 11/May/23
	$soit: C(O,R) CentreO(a,b) equatin cercle (origine O) (x−a)^2 +(y−b)^2 =R^2 (E) Points: C et B verrifient { (((4−a)^2 +(−3−b)^2 =R^2 (1))),(((3−a)^2 + (2−b)^2 =R^2 (2))) :} (2)−(1)⇒ a−5b=6 (3) distance BC=(√((4−3)^2 +(−5)^2 )) =(√(26)) le point C(x,y) a pour coordonnes △ABC AC=(√((4−x)^2 +(3+y)^2 )) ° y_A >y_C x_A <x_c AC^2 =26+(8,06)^2 −2(8,06×(√(26)) )cos (118) =116,949583656 AC=10,81432 { (),() :} (E) x^2 +y^2 −2(ax+by)=R^2 a=6+5b d apres propriete du centre cercle O 2∡BCA=∡BOA=90 ⇒△OAB triangle rectangle OA^2 +OB^2 =R^2 OA=OB=R ⇒ R=((AB(√2))/2)=5,699 d apres( E) ( x−a)^2 +(y−b)^2 =32,478 aplique au pont (BC) { (((4−a)^2 +(−3+((6−a)/5))^2 =32,478 (1))),(((3−a)^2 +(2+((6−a)/5))^2 =32,478 (2))) :} choisissons (1) a^2 −8a−6(((6−a)/5))+25+(((6−a)/5))^2 =32,478 25a^2 −200a−30(6−a)+25^2 +(6−a)^2 =32,478×25 26a^2 −182a+481=32,478×25^2 a^2 −7a−781,25=0 △=49+3125=3174=23(√6) a=((7±23(√6))/2)⇒ { ((a=((7+23(√6))/2) (1))),((a=((7−23(√6))/2) (2))) :} b=((a−6)/5) b= ((23(√6) +1)/(10)) b=((6−23(√6))/(10)) donc les coordonnes possibles du centr O sont: (a,b)={(((7+23(√6))/2) ,((23(√6) +1)/(10))) , (((7−23(√6))/2),((6−23(√6))/(10)))} equation 2⇒ 13+a^2 +b^2 −(6a+4b)=R^2 (y−b)^2 =R^2 −(x−a)^2 y=(√(R^2 −(x−a)^2 )) +b R^2 −(x−a)^2 >0 R>∣x−a∣ { ((xc=4>0)),((y_c <0)) :} ⇒ 4−((7+23(√6))/2)=((8−7−23(√6))/2)<0 ⇒y<0 donc la valeur de a acceptee est a=((7−23(√6))/2) soit b=((6−23(√6))/(10)) donc: O(((7−23(√6))/2) ,((6−23(√6))/(10)))$
	$soit : C (O, R) C entre O (a, b)$ $equatin cercle (origine O)$ ${(x - a)}^{2} + {(y - b)}^{2} = R^{2} (E)$ $Points : C et B verrifient$ ${\begin{cases} {(4 - a)}^{2} + {(- 3 - b)}^{2} = R^{2} (1) \\ {(3 - a)}^{2} + {(2 - b)}^{2} = R^{2} (2) \end{cases}$ $(2) - (1) \Rightarrow a - 5 b = 6 (3)$ $distance BC = \sqrt{{(4 - 3)}^{2} + {(- 5)}^{2}} = \sqrt{26}$ $le point C (x, y) a pour coordonnes$ $△ ABC AC = \sqrt{{(4 - x)}^{2} + {(3 + y)}^{2}}$ $° y_{A} > y_{C} x_{A} < x_{c}$ ${AC}^{2} = 26 + {(8, 06)}^{2} - 2 (8, 06 \times \sqrt{26}) \cos (118)$ $= 116, 949583656 AC = 10, 81432 {\begin{cases} \end{cases}$ $(E) x^{2} + y^{2} - 2 (ax + by) = R^{2}$ $a = 6 + 5 b$ $d a pres propriete du centre cercle O$ $2 ∡ BCA = ∡ BOA = 90$ $\Rightarrow △ OAB triangle rectangle$ ${OA}^{2} + {OB}^{2} = R^{2}$ $OA = OB = R \Rightarrow R = \frac{AB \sqrt{2}}{2} = 5, 699$ $d apres (E) {(x - a)}^{2} + {(y - b)}^{2} = 32, 478$ $aplique au pont (BC)$ ${\begin{cases} {(4 - a)}^{2} + {(- 3 + \frac{6 - a}{5})}^{2} = 32, 478 (1) \\ {(3 - a)}^{2} + {(2 + \frac{6 - a}{5})}^{2} = 32, 478 (2) \end{cases}$ $choisissons (1)$ $a^{2} - 8 a - 6 (\frac{6 - a}{5}) + 25 + {(\frac{6 - a}{5})}^{2} = 32, 478$ $25 a^{2} - 200 a - 30 (6 - a) + 25^{2} + {(6 - a)}^{2} = 32, 478 \times 25$ $26 a^{2} - 182 a + 481 = 32, 478 \times 25^{2}$ $a^{2} - 7 a - 781, 25 = 0$ $△ = 49 + 3125 = 3174 = 23 \sqrt{6}$ $a = \frac{7 \pm 23 \sqrt{6}}{2} \Rightarrow {\begin{cases} a = \frac{7 + 23 \sqrt{6}}{2} (1) \\ a = \frac{7 - 23 \sqrt{6}}{2} (2) \end{cases}$ $b = \frac{a - 6}{5} b = \frac{23 \sqrt{6} + 1}{10}$ $b = \frac{6 - 23 \sqrt{6}}{10}$ $donc les coordonnes possibles d u centr O sont :$ $(a, b) = {(\frac{7 + 23 \sqrt{6}}{2}, \frac{23 \sqrt{6} + 1}{10}), (\frac{7 - 23 \sqrt{6}}{2}, \frac{6 - 23 \sqrt{6}}{10})}$ $equation 2 \Rightarrow$ $13 + a^{2} + b^{2} - (6 a + 4 b) = R^{2}$ ${(y - b)}^{2} = R^{2} - {(x - a)}^{2}$ $y = \sqrt{R^{2} - {(x - a)}^{2}} + b$ $R^{2} - {(x - a)}^{2} > 0 R >∣ x - a ∣$ ${\begin{cases} x c = 4 > 0 \\ y_{c} < 0 \end{cases} \Rightarrow 4 - \frac{7 + 23 \sqrt{6}}{2} = \frac{8 - 7 - 23 \sqrt{6}}{2} < 0$ $\Rightarrow y < 0$ $donc la valeur de a acceptee est$ $a = \frac{7 - 23 \sqrt{6}}{2} soit b = \frac{6 - 23 \sqrt{6}}{10}$ $donc : O (\frac{7 - 23 \sqrt{6}}{2}, \frac{6 - 23 \sqrt{6}}{10})$
	Commented by naka3546 last updated on 12/May/23

	$Thank you$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum