Question Number 48966 by peter frank last updated on 30/Nov/18
Commented by tanmay.chaudhury50@gmail.com last updated on 30/Nov/18
$${purchase}\:{good}\:{books}\:{from}\:{market}..{stop}\:{reading} \\ $$$${non}\:{standard}\:{books}.. \\ $$
Commented by peter frank last updated on 30/Nov/18
$$\mathrm{thanks}\:\mathrm{sir}\:\mathrm{for}\:\mathrm{advice}\: \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 30/Nov/18
$${k}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)+{y}^{\mathrm{2}} −\mathrm{4}{x}^{\mathrm{2}} +\mathrm{3}{y}−\mathrm{6}{x}+{y}+\mathrm{2}{x}+\mathrm{3}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} \left({k}−\mathrm{4}\right)+{y}^{\mathrm{2}} \left({k}+\mathrm{1}\right)−\mathrm{4}{x}+\mathrm{4}{y}+\mathrm{3}=\mathrm{0} \\ $$$${for}\:{circle}\:{k}−\mathrm{4}={k}+\mathrm{1}\:{but}\:{it}\:{is}\:{not}\:{possible} \\ $$$${so}\:{this}\:{is}\:{not}\:{eqn}\:{of}\:{circle}...\boldsymbol{{pls}}\:\boldsymbol{{check}}\:\boldsymbol{{question}}.\mathrm{99} \\ $$