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Question Number 58254 by pete last updated on 20/Apr/19

The line 3x−2y−5=0 is parallel to a diameter  of a circle x^2 +y^2 −4x+2y−4=0. find the  equation of the diameter.

$$\mathrm{The}\:\mathrm{line}\:\mathrm{3x}−\mathrm{2y}−\mathrm{5}=\mathrm{0}\:\mathrm{is}\:\mathrm{parallel}\:\mathrm{to}\:\mathrm{a}\:\mathrm{diameter} \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{4x}+\mathrm{2y}−\mathrm{4}=\mathrm{0}.\:\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{diameter}. \\ $$

Answered by tanmay last updated on 20/Apr/19

x^2 +y^2 −4x+2y−4=0  x^2 −4x+4+y^2 +2y+1−5−4=0  (x−2)^2 +(y+1)^2 =3^2   centre of circle (2,−1) radius=3  3x−2y−5=0  2y=3x−5  y=(3/2)x−(5/2)→slope m=(3/2)  eqn diameter  (y+1)=(3/2)(x−2)  2y+2=3x−6  3x−2y−8=0

$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{2}{y}−\mathrm{4}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4}+{y}^{\mathrm{2}} +\mathrm{2}{y}+\mathrm{1}−\mathrm{5}−\mathrm{4}=\mathrm{0} \\ $$$$\left({x}−\mathrm{2}\right)^{\mathrm{2}} +\left({y}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{3}^{\mathrm{2}} \\ $$$${centre}\:{of}\:{circle}\:\left(\mathrm{2},−\mathrm{1}\right)\:{radius}=\mathrm{3} \\ $$$$\mathrm{3}{x}−\mathrm{2}{y}−\mathrm{5}=\mathrm{0} \\ $$$$\mathrm{2}{y}=\mathrm{3}{x}−\mathrm{5} \\ $$$${y}=\frac{\mathrm{3}}{\mathrm{2}}{x}−\frac{\mathrm{5}}{\mathrm{2}}\rightarrow{slope}\:{m}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${eqn}\:{diameter} \\ $$$$\left({y}+\mathrm{1}\right)=\frac{\mathrm{3}}{\mathrm{2}}\left({x}−\mathrm{2}\right) \\ $$$$\mathrm{2}{y}+\mathrm{2}=\mathrm{3}{x}−\mathrm{6} \\ $$$$\mathrm{3}{x}−\mathrm{2}{y}−\mathrm{8}=\mathrm{0} \\ $$

Commented by pete last updated on 20/Apr/19

Thanks very much sir

$$\mathrm{Thanks}\:\mathrm{very}\:\mathrm{much}\:\mathrm{sir} \\ $$

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