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Question Number 48966 by peter frank last updated on 30/Nov/18

Commented by tanmay.chaudhury50@gmail.com last updated on 30/Nov/18

$$purchasegoodbooksfrommarket..stopreading$$$$nonstandardbooks..$$

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

$$k(x^2+y^2)+y^2-4x^2+3y-6x+y+2x+3=0$$$$x^2(k-4)+y^2(k+1)-4x+4y+3=0$$$$forcirclek-4=k+1butitisnotpossible$$$$sothisisnoteqnofcircle...\mathit{p}\mathit{l}\mathit{s}\mathit{c}\mathit{h}\mathit{e}\mathit{c}\mathit{k}\mathit{q}\mathit{u}\mathit{e}\mathit{s}\mathit{t}\mathit{i}\mathit{o}\mathit{n}.99$$

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