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Question Number 33511 by mondodotto@gmail.com last updated on 18/Apr/18

$$\mathbf{find}\mathbf{the}\mathbf{value}\mathbf{of}{}^{\prime }{\mathbf{k}}^{\prime }\mathbf{such}\mathbf{that}$$$$\mathbf{k}({\mathit{x}}^2+{\mathit{y}}^2)+(\mathit{y}-2\mathit{x}+1)(\mathit{y}+2\mathit{x}+3)=0$$$$\mathbf{is}\mathbf{a}\mathbf{circle}.\mathbf{Hence}\mathbf{obtain}\mathbf{the}$$$$\mathbf{centre}\mathbf{and}\mathbf{radius}\mathbf{of}\mathbf{the}\mathbf{resulting}\mathbf{circle}.$$

Answered by MJS last updated on 18/Apr/18

$$(k-4)x^2+(k+1)y^2-4x+4y+3=0$$$$$$$$\mathrm{any}\mathrm{circle}\mathrm{has}\mathrm{the}\mathrm{equation}$$$${(x-m)}^2+{(y-n)}^2-r^2=0$$$$\mathrm{you}\mathrm{can}\mathrm{multiply}\mathrm{this}\mathrm{with}\mathrm{a}$$$$c\mathrm{onstant}$$$$c\times {(x-m)}^2+c\times {(y-n)}^2-c\times r^2=0$$$$\mathrm{but}\mathrm{the}\mathrm{coefficients}\mathrm{of}{x}^2\mathrm{and}{y}^2$$$$\mathrm{are}\mathrm{always}\mathrm{identical}$$$$$$$$k-4=k+1$$$$\mathrm{has}\mathrm{no}\mathrm{solution}\Rightarrow \mathrm{no}\mathrm{circle}$$

Commented by mondodotto@gmail.com last updated on 18/Apr/18

$$\mathrm{thanx}\mathrm{a}\mathrm{lot}\mathrm{god}\mathrm{bless}\mathrm{you}$$

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