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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
write the equation of the circle which containing the points (4,−3),(1,−2) and center on line 3x + 4y =7
$${write}\:{the}\:{equation}\:{of}\:{the}\: \\ $$$${circle}\:{which}\:{containing}\: \\ $$$${the}\:{points}\:\left(\mathrm{4},−\mathrm{3}\right),\left(\mathrm{1},−\mathrm{2}\right) \\ $$$${and}\:{center}\:{on}\:{line} \\ $$$$\mathrm{3}{x}\:+\:\mathrm{4}{y}\:=\mathrm{7} \\ $$
Answered by mr W last updated on 09/Jan/21
(((4+1)/2), ((−3−2)/2))=((5/2), −(5/2)) ((−3−(−2))/(4−1))=−(1/3) y=−(5/2)+3(x−(5/2)) −(5/2)+3(x−(5/2))=(7/4)−(3/4)x ⇒x=((47)/(15)) ⇒y=−(5/2)+3(((47)/(15))−(5/2))=−(3/5) ⇒center (((47)/(15)), −(3/5)) radius^2 =(4−((47)/(15)))^2 +(−3+(3/5))^2 =((293)/(45)) ⇒eqn. of circle: (x−((47)/(15)))^2 +(y+(3/5))^2 =((293)/(45))
$$\left(\frac{\mathrm{4}+\mathrm{1}}{\mathrm{2}},\:\frac{−\mathrm{3}−\mathrm{2}}{\mathrm{2}}\right)=\left(\frac{\mathrm{5}}{\mathrm{2}},\:−\frac{\mathrm{5}}{\mathrm{2}}\right) \\ $$$$\frac{−\mathrm{3}−\left(−\mathrm{2}\right)}{\mathrm{4}−\mathrm{1}}=−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${y}=−\frac{\mathrm{5}}{\mathrm{2}}+\mathrm{3}\left({x}−\frac{\mathrm{5}}{\mathrm{2}}\right) \\ $$$$−\frac{\mathrm{5}}{\mathrm{2}}+\mathrm{3}\left({x}−\frac{\mathrm{5}}{\mathrm{2}}\right)=\frac{\mathrm{7}}{\mathrm{4}}−\frac{\mathrm{3}}{\mathrm{4}}{x} \\ $$$$\Rightarrow{x}=\frac{\mathrm{47}}{\mathrm{15}} \\ $$$$\Rightarrow{y}=−\frac{\mathrm{5}}{\mathrm{2}}+\mathrm{3}\left(\frac{\mathrm{47}}{\mathrm{15}}−\frac{\mathrm{5}}{\mathrm{2}}\right)=−\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\Rightarrow{center}\:\left(\frac{\mathrm{47}}{\mathrm{15}},\:−\frac{\mathrm{3}}{\mathrm{5}}\right) \\ $$$${radius}^{\mathrm{2}} =\left(\mathrm{4}−\frac{\mathrm{47}}{\mathrm{15}}\right)^{\mathrm{2}} +\left(−\mathrm{3}+\frac{\mathrm{3}}{\mathrm{5}}\right)^{\mathrm{2}} =\frac{\mathrm{293}}{\mathrm{45}} \\ $$$$\Rightarrow{eqn}.\:{of}\:{circle}: \\ $$$$\left({x}−\frac{\mathrm{47}}{\mathrm{15}}\right)^{\mathrm{2}} +\left({y}+\frac{\mathrm{3}}{\mathrm{5}}\right)^{\mathrm{2}} =\frac{\mathrm{293}}{\mathrm{45}} \\ $$
Commented by malwan last updated on 09/Jan/21
thank you sir
$${thank}\:{you}\:{sir} \\ $$
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))
$$\mathrm{let}\:\mathrm{the}\:\mathrm{center}\:\mathrm{point}\:\mathrm{is}\:\mathrm{P}\left(\mathrm{a},\mathrm{b}\right)\: \\ $$$$\left(\mathrm{1}\right)\:\mathrm{R}=\sqrt{\left(\mathrm{4}−\mathrm{a}\right)^{\mathrm{2}} +\left(−\mathrm{3}−\mathrm{b}\right)^{\mathrm{2}} } \\ $$$$\left(\mathrm{2}\right)\mathrm{R}=\sqrt{\left(\mathrm{1}−\mathrm{a}\right)^{\mathrm{2}} +\left(−\mathrm{2}−\mathrm{b}\right)^{\mathrm{2}} } \\ $$$$\left(\mathrm{3}\right)\:\mathrm{R}=\mathrm{R}\:\Rightarrow\mathrm{16}−\mathrm{8a}+\mathrm{a}^{\mathrm{2}} +\mathrm{9}+\mathrm{6b}+\mathrm{b}^{\mathrm{2}} =\mathrm{1}−\mathrm{2a}+\mathrm{a}^{\mathrm{2}} +\mathrm{4}+\mathrm{4b}+\mathrm{b}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{25}−\mathrm{8a}+\mathrm{6b}=\mathrm{5}−\mathrm{2a}+\mathrm{4b}\:;\:−\mathrm{6a}+\mathrm{2b}=−\mathrm{20} \\ $$$$\:\:\:\:\:\:−\mathrm{3a}+\mathrm{b}=−\mathrm{10} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{3a}+\mathrm{4b}=\mathrm{7}\: \\ $$$$\left(\mathrm{2}\right)+\left(\mathrm{3}\right)\Rightarrow\mathrm{5b}=−\mathrm{3}\:,\:\begin{cases}{\mathrm{b}=−\frac{\mathrm{3}}{\mathrm{5}}}\\{\mathrm{a}=\frac{\mathrm{7}−\mathrm{4}\left(−\frac{\mathrm{3}}{\mathrm{5}}\right)}{\mathrm{3}}=\frac{\mathrm{47}}{\mathrm{15}}}\end{cases} \\ $$$$ \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{P}\left(\frac{\mathrm{47}}{\mathrm{15}},\:−\frac{\mathrm{3}}{\mathrm{5}}\right)\:\&\:\mathrm{R}=\sqrt{\left(\mathrm{4}−\frac{\mathrm{47}}{\mathrm{15}}\right)^{\mathrm{2}} +\left(−\mathrm{3}+\frac{\mathrm{3}}{\mathrm{5}}\right)^{\mathrm{2}} }=\sqrt{\left(\frac{\mathrm{169}}{\mathrm{225}}\right)+\left(\frac{\mathrm{144}}{\mathrm{25}}\right)} \\ $$$$\mathrm{R}=\sqrt{\frac{\mathrm{1465}}{\mathrm{225}}}\:.\:\mathrm{then}\:\mathrm{eq}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\::\equiv\:\left(\mathrm{x}−\frac{\mathrm{47}}{\mathrm{15}}\right)^{\mathrm{2}} +\left(\mathrm{y}+\frac{\mathrm{3}}{\mathrm{5}}\right)^{\mathrm{2}} =\frac{\mathrm{1465}}{\mathrm{225}} \\ $$
Commented by malwan last updated on 09/Jan/21
thank you so much
$${thank}\:{you}\:{so}\:{much} \\ $$

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