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Question Number 1012 by rpatle69@gmail.com last updated on 13/May/15

find the equation of circle whose  parametric form is given by   x=3cos θ+5 and y= 3sin θ−7 and  second part is x=4cos θ−3 and  y=4sin θ+4. find centre and   radius of above circle.      guys plz ans. me soon.....

$${find}\:{the}\:{equation}\:{of}\:{circle}\:{whose} \\ $$$${parametric}\:{form}\:{is}\:{given}\:{by}\: \\ $$$${x}=\mathrm{3cos}\:\theta+\mathrm{5}\:{and}\:{y}=\:\mathrm{3sin}\:\theta−\mathrm{7}\:{and} \\ $$$${second}\:{part}\:{is}\:{x}=\mathrm{4cos}\:\theta−\mathrm{3}\:{and} \\ $$$${y}=\mathrm{4sin}\:\theta+\mathrm{4}.\:{find}\:{centre}\:{and}\: \\ $$$${radius}\:{of}\:{above}\:{circle}. \\ $$$$ \\ $$$$ \\ $$$${guys}\:{plz}\:{ans}.\:{me}\:{soon}..... \\ $$$$ \\ $$

Answered by sudhanshur last updated on 13/May/15

x−5=3cos θ  y+7=3sin θ  square and add  (x−5)^2 +(y+7)^2 =9

$${x}−\mathrm{5}=\mathrm{3cos}\:\theta \\ $$$${y}+\mathrm{7}=\mathrm{3sin}\:\theta \\ $$$${square}\:{and}\:{add} \\ $$$$\left({x}−\mathrm{5}\right)^{\mathrm{2}} +\left({y}+\mathrm{7}\right)^{\mathrm{2}} =\mathrm{9} \\ $$

Answered by sudhanshur last updated on 13/May/15

x+3=4cos θ  y−4=4sin θ  (x+3)^2 +(y−4)^2 =16

$${x}+\mathrm{3}=\mathrm{4cos}\:\theta \\ $$$${y}−\mathrm{4}=\mathrm{4sin}\:\theta \\ $$$$\left({x}+\mathrm{3}\right)^{\mathrm{2}} +\left({y}−\mathrm{4}\right)^{\mathrm{2}} =\mathrm{16} \\ $$

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