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Question Number 140597 by bramlexs22 last updated on 10/May/21
Find the equation of circle which passes through the point (2,0) and whose center is the limit of the point of intersection of the lines 3x+5y=1 and (2+c)x+5c^2 y=1 as c→1 .
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{circle}\:\mathrm{which} \\ $$$$\mathrm{passes}\:\mathrm{through}\:\mathrm{the}\:\mathrm{point}\:\left(\mathrm{2},\mathrm{0}\right)\:\mathrm{and} \\ $$$$\mathrm{whose}\:\mathrm{center}\:\mathrm{is}\:\mathrm{the}\:\mathrm{limit}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{point}\:\mathrm{of}\:\mathrm{intersection}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{lines}\:\mathrm{3x}+\mathrm{5y}=\mathrm{1}\:\mathrm{and}\:\left(\mathrm{2}+\mathrm{c}\right)\mathrm{x}+\mathrm{5c}^{\mathrm{2}} \mathrm{y}=\mathrm{1} \\ $$$$\mathrm{as}\:\mathrm{c}\rightarrow\mathrm{1}\:. \\ $$
Answered by TheSupreme last updated on 10/May/21
3x+5y=1 (2+c)x+5c^2 y=1 (c−1)x+5(c^2 −1)y=0 y=(x/(5(c+1))) 3x+(x/(c+1))=1 → x(((4c+3)/(c+1)))=1 x=((c+1)/(4c+3))→(2/7) y→ (1/(35)) (x−(2/7))^2 +(y−(1/(35)))^2 =R^2 (2,0) (((12)/7))^2 +((1/(35)))^2 =R^2 (1/7^2 )(144+(1/(25)))=R^2 ((361)/(35^2 )) =R^2 (x−(2/7))^2 +(y−(1/(35)))^2 =((361)/(35^2 ))
$$\mathrm{3}{x}+\mathrm{5}{y}=\mathrm{1} \\ $$$$\left(\mathrm{2}+{c}\right){x}+\mathrm{5}{c}^{\mathrm{2}} {y}=\mathrm{1} \\ $$$$\left({c}−\mathrm{1}\right){x}+\mathrm{5}\left({c}^{\mathrm{2}} −\mathrm{1}\right){y}=\mathrm{0} \\ $$$${y}=\frac{{x}}{\mathrm{5}\left({c}+\mathrm{1}\right)} \\ $$$$\mathrm{3}{x}+\frac{{x}}{{c}+\mathrm{1}}=\mathrm{1}\:\rightarrow\:{x}\left(\frac{\mathrm{4}{c}+\mathrm{3}}{{c}+\mathrm{1}}\right)=\mathrm{1} \\ $$$${x}=\frac{{c}+\mathrm{1}}{\mathrm{4}{c}+\mathrm{3}}\rightarrow\frac{\mathrm{2}}{\mathrm{7}} \\ $$$${y}\rightarrow\:\frac{\mathrm{1}}{\mathrm{35}} \\ $$$$\left({x}−\frac{\mathrm{2}}{\mathrm{7}}\right)^{\mathrm{2}} +\left({y}−\frac{\mathrm{1}}{\mathrm{35}}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\left(\mathrm{2},\mathrm{0}\right) \\ $$$$\left(\frac{\mathrm{12}}{\mathrm{7}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{35}}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{7}^{\mathrm{2}} }\left(\mathrm{144}+\frac{\mathrm{1}}{\mathrm{25}}\right)={R}^{\mathrm{2}} \\ $$$$\frac{\mathrm{361}}{\mathrm{35}^{\mathrm{2}} }\:={R}^{\mathrm{2}} \\ $$$$\left({x}−\frac{\mathrm{2}}{\mathrm{7}}\right)^{\mathrm{2}} +\left({y}−\frac{\mathrm{1}}{\mathrm{35}}\right)^{\mathrm{2}} =\frac{\mathrm{361}}{\mathrm{35}^{\mathrm{2}} } \\ $$
Commented by bramlexs22 last updated on 10/May/21
yes. thank you
$$\mathrm{yes}.\:\mathrm{thank}\:\mathrm{you} \\ $$

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