write the equation of the circle which containing the points (4,−3),(1,−2) and center on line 3x + 4y =7


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Question Number 128611 by malwan last updated on 08/Jan/21

$w r i t e t h e e q u a t i o n o f t h e$ $c i r c l e w h i c h c o n t a i n i n g$ $t h e p o i n t s (4, - 3), (1, - 2)$ $a n d c e n t e r o n l i n e$ $3 x + 4 y = 7$
	Answered by mr W last updated on 09/Jan/21

	$(\frac{4 + 1}{2}, \frac{- 3 - 2}{2}) = (\frac{5}{2}, - \frac{5}{2})$ $\frac{- 3 - (- 2)}{4 - 1} = - \frac{1}{3}$ $y = - \frac{5}{2} + 3 (x - \frac{5}{2})$ $- \frac{5}{2} + 3 (x - \frac{5}{2}) = \frac{7}{4} - \frac{3}{4} x$ $\Rightarrow x = \frac{47}{15}$ $\Rightarrow y = - \frac{5}{2} + 3 (\frac{47}{15} - \frac{5}{2}) = - \frac{3}{5}$ $\Rightarrow c e n t e r (\frac{47}{15}, - \frac{3}{5})$ ${r a d i u s}^{2} = {(4 - \frac{47}{15})}^{2} + {(- 3 + \frac{3}{5})}^{2} = \frac{293}{45}$ $\Rightarrow e q n . o f c i r c l e :$ ${(x - \frac{47}{15})}^{2} + {(y + \frac{3}{5})}^{2} = \frac{293}{45}$
	Commented by malwan last updated on 09/Jan/21

	$t h a n k y o u s i r$
	Answered by liberty last updated on 08/Jan/21
	$let the center point is P(a,b) (1) R=(√((4−a)^2 +(−3−b)^2 )) (2)R=(√((1−a)^2 +(−2−b)^2 )) (3) R=R ⇒16−8a+a^2 +9+6b+b^2 =1−2a+a^2 +4+4b+b^2 ⇒25−8a+6b=5−2a+4b ; −6a+2b=−20 −3a+b=−10 (3) 3a+4b=7 (2)+(3)⇒5b=−3 , { ((b=−(3/5))),((a=((7−4(−(3/5)))/3)=((47)/(15)))) :} we get P(((47)/(15)), −(3/5)) & R=(√((4−((47)/(15)))^2 +(−3+(3/5))^2 ))=(√((((169)/(225)))+(((144)/(25))))) R=(√((1465)/(225))) . then eq of the circle :≡ (x−((47)/(15)))^2 +(y+(3/5))^2 =((1465)/(225))$
	$let the center point is P (a, b)$ $(1) R = \sqrt{{(4 - a)}^{2} + {(- 3 - b)}^{2}}$ $(2) R = \sqrt{{(1 - a)}^{2} + {(- 2 - b)}^{2}}$ $(3) R = R \Rightarrow 16 - 8 a + a^{2} + 9 + 6 b + b^{2} = 1 - 2 a + a^{2} + 4 + 4 b + b^{2}$ $\Rightarrow 25 - 8 a + 6 b = 5 - 2 a + 4 b; - 6 a + 2 b = - 20$ $- 3 a + b = - 10$ $(3) 3 a + 4 b = 7$ $(2) + (3) \Rightarrow 5 b = - 3, {\begin{cases} b = - \frac{3}{5} \\ a = \frac{7 - 4 (- \frac{3}{5})}{3} = \frac{47}{15} \end{cases}$ $we get P (\frac{47}{15}, - \frac{3}{5}) & R = \sqrt{{(4 - \frac{47}{15})}^{2} + {(- 3 + \frac{3}{5})}^{2}} = \sqrt{(\frac{169}{225}) + (\frac{144}{25})}$ $R = \sqrt{\frac{1465}{225}} . then eq of the circle :\equiv {(x - \frac{47}{15})}^{2} + {(y + \frac{3}{5})}^{2} = \frac{1465}{225}$
	Commented by malwan last updated on 09/Jan/21

	$t h a n k y o u s o m u c h$
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