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Question Number 222084 by atara last updated on 17/Jun/25
find the equation of the cirvle 3x^2 +3y^2 −12x−6x−45>0    x^2 +y^2 −4x−2y−15>0  x^2 −4x+y^2 −2y>15  x^2 −4x+((1/2)×4)^2 +y^2 −2y+((1/2)×2)^2 >15+4+1  x^2 −4x+4+y^2 −2y+1>20    (x−2)^2 +(x−1)^2 >20  (x−h)^2 +(y−k)^2 =r^2   c(h      a
$${find}\:{the}\:{equation}\:{of}\:{the}\:{cirvle}\:\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{y}^{\mathrm{2}} −\mathrm{12}{x}−\mathrm{6}{x}−\mathrm{45}>\mathrm{0} \\ $$$$ \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{2}{y}−\mathrm{15}>\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\mathrm{4}{x}+{y}^{\mathrm{2}} −\mathrm{2}{y}>\mathrm{15} \\ $$$${x}^{\mathrm{2}} −\mathrm{4}{x}+\left(\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{4}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{y}+\left(\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}\right)^{\mathrm{2}} >\mathrm{15}+\mathrm{4}+\mathrm{1} \\ $$$${x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4}+{y}^{\mathrm{2}} −\mathrm{2}{y}+\mathrm{1}>\mathrm{20} \\ $$$$ \\ $$$$\left({x}−\mathrm{2}\right)^{\mathrm{2}} +\left({x}−\mathrm{1}\right)^{\mathrm{2}} >\mathrm{20} \\ $$$$\left({x}−{h}\right)^{\mathrm{2}} +\left({y}−{k}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${c}\left({h}\right. \\ $$$$ \\ $$$$ \\ $$$${a} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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