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Question Number 33511 by mondodotto@gmail.com last updated on 18/Apr/18
find the value of ′k′ such that  k(x^2 +y^2 )+(y−2x+1)(y+2x+3)=0  is a circle.Hence obtain the  centre and radius of the resulting circle.
$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\:'\boldsymbol{\mathrm{k}}'\:\boldsymbol{\mathrm{such}}\:\boldsymbol{\mathrm{that}} \\ $$$$\boldsymbol{\mathrm{k}}\left(\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} \right)+\left(\boldsymbol{{y}}−\mathrm{2}\boldsymbol{{x}}+\mathrm{1}\right)\left(\boldsymbol{{y}}+\mathrm{2}\boldsymbol{{x}}+\mathrm{3}\right)=\mathrm{0} \\ $$$$\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{circle}}.\boldsymbol{\mathrm{Hence}}\:\boldsymbol{\mathrm{obtain}}\:\boldsymbol{\mathrm{the}} \\ $$$$\boldsymbol{\mathrm{centre}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{radius}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{resulting}}\:\boldsymbol{\mathrm{circle}}. \\ $$
Answered by MJS last updated on 18/Apr/18
(k−4)x^2 +(k+1)y^2 −4x+4y+3=0    any circle has the equation  (x−m)^2 +(y−n)^2 −r^2 =0  you can multiply this with a  constant  c×(x−m)^2 +c×(y−n)^2 −c×r^2 =0  but the coefficients of x^2  and y^2   are always identical    k−4=k+1  has no solution ⇒ no circle
$$\left({k}−\mathrm{4}\right){x}^{\mathrm{2}} +\left({k}+\mathrm{1}\right){y}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4}{y}+\mathrm{3}=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{any}\:\mathrm{circle}\:\mathrm{has}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\left({x}−{m}\right)^{\mathrm{2}} +\left({y}−{n}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{you}\:\mathrm{can}\:\mathrm{multiply}\:\mathrm{this}\:\mathrm{with}\:\mathrm{a} \\ $$$${c}\mathrm{onstant} \\ $$$${c}×\left({x}−{m}\right)^{\mathrm{2}} +{c}×\left({y}−{n}\right)^{\mathrm{2}} −{c}×{r}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{but}\:\mathrm{the}\:\mathrm{coefficients}\:\mathrm{of}\:{x}^{\mathrm{2}} \:\mathrm{and}\:{y}^{\mathrm{2}} \\ $$$$\mathrm{are}\:\mathrm{always}\:\mathrm{identical} \\ $$$$ \\ $$$${k}−\mathrm{4}={k}+\mathrm{1} \\ $$$$\mathrm{has}\:\mathrm{no}\:\mathrm{solution}\:\Rightarrow\:\mathrm{no}\:\mathrm{circle} \\ $$
Commented by mondodotto@gmail.com last updated on 18/Apr/18
thanx a lot god bless you
$$\mathrm{thanx}\:\mathrm{a}\:\mathrm{lot}\:\mathrm{god}\:\mathrm{bless}\:\mathrm{you} \\ $$

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