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Question Number 10660 by okhema last updated on 22/Feb/17

ind the centre and radius of these:  (a) 6x^2 +6y^2 −4x−5y−2=0  (b) x^2 +y^2 +6x+8y−1=0  (c) 3x^2 +3y^2 −4x+8y−2=0

$${ind}\:{the}\:{centre}\:{and}\:{radius}\:{of}\:{these}: \\ $$$$\left({a}\right)\:\mathrm{6}{x}^{\mathrm{2}} +\mathrm{6}{y}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{5}{y}−\mathrm{2}=\mathrm{0} \\ $$$$\left({b}\right)\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{8}{y}−\mathrm{1}=\mathrm{0} \\ $$$$\left({c}\right)\:\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{y}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{8}{y}−\mathrm{2}=\mathrm{0} \\ $$

Answered by sandy_suhendra last updated on 22/Feb/17

a) 6x^2 +6y^2 −4x−5y−2=0          x^2 +y^2 −(2/3)x−(5/6)y=(1/3)        (x^2 −(2/3)x+(1/9))+(y^2 −(5/6)y+((25)/(144)))=(1/3)+(1/9)+((25)/(144))             (x−(1/3))^2 + (y−(5/(12)))^2 = ((89)/(144))      the centre : ((1/3) , (5/(12)))  the radius=(√((89)/(144))) = (1/(12))(√(89))

$$\left.\mathrm{a}\right)\:\mathrm{6x}^{\mathrm{2}} +\mathrm{6y}^{\mathrm{2}} −\mathrm{4x}−\mathrm{5y}−\mathrm{2}=\mathrm{0}\: \\ $$$$\:\:\:\:\:\:\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\frac{\mathrm{2}}{\mathrm{3}}\mathrm{x}−\frac{\mathrm{5}}{\mathrm{6}}\mathrm{y}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\left(\mathrm{x}^{\mathrm{2}} −\frac{\mathrm{2}}{\mathrm{3}}\mathrm{x}+\frac{\mathrm{1}}{\mathrm{9}}\right)+\left(\mathrm{y}^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{6}}\mathrm{y}+\frac{\mathrm{25}}{\mathrm{144}}\right)=\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{9}}+\frac{\mathrm{25}}{\mathrm{144}}\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} +\:\left(\mathrm{y}−\frac{\mathrm{5}}{\mathrm{12}}\right)^{\mathrm{2}} =\:\frac{\mathrm{89}}{\mathrm{144}}\:\:\:\: \\ $$$$\mathrm{the}\:\mathrm{centre}\::\:\left(\frac{\mathrm{1}}{\mathrm{3}}\:,\:\frac{\mathrm{5}}{\mathrm{12}}\right) \\ $$$$\mathrm{the}\:\mathrm{radius}=\sqrt{\frac{\mathrm{89}}{\mathrm{144}}}\:=\:\frac{\mathrm{1}}{\mathrm{12}}\sqrt{\mathrm{89}} \\ $$

Answered by sandy_suhendra last updated on 22/Feb/17

b) x^2 +6x+y^2 +8y=1      (x^2 +6x+9)+(y^2 +8y+16)=1+9+16          (x+3)^2 +(y+4)^2 =26  the centre : (−3,−4)  the radius=(√(26))    c) 3x^2 −4x+3y^2 +8y=2       x^2 −(4/3)x+y^2 +(8/3)y=(2/3)     (x^2 −(4/3)x+(4/9))+(y^2 +(8/3)y+((16)/9))=(2/3)+(4/9)+((16)/9)         (x^2 −(2/3))^2 +(y+(4/3))^2 =((26)/9)  the centre : ((2/3) , −(4/3))  the radius = (√((26)/9)) = (1/3)(√(26))

$$\left.\mathrm{b}\right)\:\mathrm{x}^{\mathrm{2}} +\mathrm{6x}+\mathrm{y}^{\mathrm{2}} +\mathrm{8y}=\mathrm{1} \\ $$$$\:\:\:\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{6x}+\mathrm{9}\right)+\left(\mathrm{y}^{\mathrm{2}} +\mathrm{8y}+\mathrm{16}\right)=\mathrm{1}+\mathrm{9}+\mathrm{16}\:\:\: \\ $$$$\:\:\:\:\:\left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{2}} +\left(\mathrm{y}+\mathrm{4}\right)^{\mathrm{2}} =\mathrm{26} \\ $$$$\mathrm{the}\:\mathrm{centre}\::\:\left(−\mathrm{3},−\mathrm{4}\right) \\ $$$$\mathrm{the}\:\mathrm{radius}=\sqrt{\mathrm{26}} \\ $$$$ \\ $$$$\left.\mathrm{c}\right)\:\mathrm{3x}^{\mathrm{2}} −\mathrm{4x}+\mathrm{3y}^{\mathrm{2}} +\mathrm{8y}=\mathrm{2} \\ $$$$\:\:\:\:\:\mathrm{x}^{\mathrm{2}} −\frac{\mathrm{4}}{\mathrm{3}}\mathrm{x}+\mathrm{y}^{\mathrm{2}} +\frac{\mathrm{8}}{\mathrm{3}}\mathrm{y}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\:\:\:\left(\mathrm{x}^{\mathrm{2}} −\frac{\mathrm{4}}{\mathrm{3}}\mathrm{x}+\frac{\mathrm{4}}{\mathrm{9}}\right)+\left(\mathrm{y}^{\mathrm{2}} +\frac{\mathrm{8}}{\mathrm{3}}\mathrm{y}+\frac{\mathrm{16}}{\mathrm{9}}\right)=\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{4}}{\mathrm{9}}+\frac{\mathrm{16}}{\mathrm{9}}\:\:\: \\ $$$$\:\:\:\:\left(\mathrm{x}^{\mathrm{2}} −\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} +\left(\mathrm{y}+\frac{\mathrm{4}}{\mathrm{3}}\right)^{\mathrm{2}} =\frac{\mathrm{26}}{\mathrm{9}} \\ $$$$\mathrm{the}\:\mathrm{centre}\::\:\left(\frac{\mathrm{2}}{\mathrm{3}}\:,\:−\frac{\mathrm{4}}{\mathrm{3}}\right) \\ $$$$\mathrm{the}\:\mathrm{radius}\:=\:\sqrt{\frac{\mathrm{26}}{\mathrm{9}}}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\mathrm{26}} \\ $$

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