Menu Close

if-a-circle-having-an-equation-x-2-y-2-6x-8y-0-is-intersected-at-A-and-B-by-x-y-1-find-the-equation-of-the-circle-on-AB-as-diameter-

Tinku Tara 0 Comments
Pin




Question Number 84157 by redmiiuser last updated on 10/Mar/20
if a circle having an equation x^2 +y^2 −6x−8y=0 is intersected at A and B by x+y=1.find the equation of the circle on AB as diameter
$${if}\:{a}\:{circle}\:{having}\:{an}\: \\ $$$${equation}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{6}{x}−\mathrm{8}{y}=\mathrm{0} \\ $$$${is}\:{intersected}\:{at}\:{A}\:{and}\:{B} \\ $$$${by}\:{x}+{y}=\mathrm{1}.{find}\:{the}\: \\ $$$${equation}\:{of}\:{the}\:{circle} \\ $$$${on}\:{AB}\:{as}\:{diameter} \\ $$
Answered by TANMAY PANACEA last updated on 10/Mar/20
x^2 +(1−x)^2 −6x−8(1−x)=0 x^2 +1−2x+x^2 −6x−8+8x=0 2x^2 −7=0 x=±(√(7/2)) A((√(7/2)) ,1−(√(7/2)) ) B(−(√(7/2)) ,1+(√(7/2)) ) centre of required circle mid point of AB (0,1) diameter=distance between AB 2r=[(2(√(7/2)) )^2 +(2(√(7/2)) )^2 ]^(1/2) 2r=[(4×(7/2)+4×(7/2))]^(1/2) =2(√7) r=(√7) eqn circle (x−0)^2 +(y−1)^2 =((√7) )^2 x^2 +y^2 −2y−6=0
$${x}^{\mathrm{2}} +\left(\mathrm{1}−{x}\right)^{\mathrm{2}} −\mathrm{6}{x}−\mathrm{8}\left(\mathrm{1}−{x}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}{x}+{x}^{\mathrm{2}} −\mathrm{6}{x}−\mathrm{8}+\mathrm{8}{x}=\mathrm{0} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{7}=\mathrm{0}\:\:\:\:{x}=\pm\sqrt{\frac{\mathrm{7}}{\mathrm{2}}}\: \\ $$$${A}\left(\sqrt{\frac{\mathrm{7}}{\mathrm{2}}}\:,\mathrm{1}−\sqrt{\frac{\mathrm{7}}{\mathrm{2}}}\:\right)\:\:{B}\left(−\sqrt{\frac{\mathrm{7}}{\mathrm{2}}}\:,\mathrm{1}+\sqrt{\frac{\mathrm{7}}{\mathrm{2}}}\:\right) \\ $$$${centre}\:{of}\:{required}\:{circle}\:{mid}\:{point}\:{of}\:{AB} \\ $$$$\left(\mathrm{0},\mathrm{1}\right)\:\:{diameter}={distance}\:{between}\:{AB} \\ $$$$\mathrm{2}{r}=\left[\left(\mathrm{2}\sqrt{\frac{\mathrm{7}}{\mathrm{2}}}\:\right)^{\mathrm{2}} +\left(\mathrm{2}\sqrt{\frac{\mathrm{7}}{\mathrm{2}}}\:\right)^{\mathrm{2}} \right]^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\mathrm{2}{r}=\left[\left(\mathrm{4}×\frac{\mathrm{7}}{\mathrm{2}}+\mathrm{4}×\frac{\mathrm{7}}{\mathrm{2}}\right)\right]^{\frac{\mathrm{1}}{\mathrm{2}}} =\mathrm{2}\sqrt{\mathrm{7}}\: \\ $$$${r}=\sqrt{\mathrm{7}}\: \\ $$$${eqn}\:{circle} \\ $$$$\left({x}−\mathrm{0}\right)^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} =\left(\sqrt{\mathrm{7}}\:\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{y}−\mathrm{6}=\mathrm{0} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Answered by john santu last updated on 10/Mar/20
the equation of circle (x−(√(7/2)))(x+(√(7/2)))+(y−(1+(√(7/2))))(y−(1−(√(7/2))))=0 x^2 −(7/2)+y^2 −2y−(5/2) = 0 x^2 + y^2 −2y − 6 =0
$$\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{circle}\: \\ $$$$\left(\mathrm{x}−\sqrt{\frac{\mathrm{7}}{\mathrm{2}}}\right)\left(\mathrm{x}+\sqrt{\frac{\mathrm{7}}{\mathrm{2}}}\right)+\left(\mathrm{y}−\left(\mathrm{1}+\sqrt{\frac{\mathrm{7}}{\mathrm{2}}}\right)\right)\left(\mathrm{y}−\left(\mathrm{1}−\sqrt{\frac{\mathrm{7}}{\mathrm{2}}}\right)\right)=\mathrm{0} \\ $$$$\mathrm{x}^{\mathrm{2}} −\frac{\mathrm{7}}{\mathrm{2}}+\mathrm{y}^{\mathrm{2}} −\mathrm{2y}−\frac{\mathrm{5}}{\mathrm{2}}\:=\:\mathrm{0} \\ $$$$\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \:−\mathrm{2y}\:−\:\mathrm{6}\:=\mathrm{0} \\ $$
Answered by TANMAY PANACEA last updated on 10/Mar/20
another way the eqn of circle (x^2 +y^2 −6x−8y)+λ(x+y−1)=0 x^2 +y^2 +x(−6+λ)+y(−8+λ)−λ=0 centre of required cirle is comparing with x^2 +y^2 +2gx+2fy+c=0 2gx=(−6+λ)x→2g=−6+λ g=(((λ−6)/2)) 2fy=(−8+λ)y→2f=−8+λ f=(((λ−8)/2)) centre of required circle(−g,−f) (((6−λ)/2),((8−λ)/2)) which lies on x+y=1 so ((6−λ)/2)+((8−λ)/2)=1→7−λ=1 so λ=6 reqired circle x^2 +y^2 +x(−6+6)+y(−8+6)−6=0 x^2 +y^2 −2y−6=0
$${another}\:{way} \\ $$$${the}\:{eqn}\:{of}\:{circle} \\ $$$$\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{6}{x}−\mathrm{8}{y}\right)+\lambda\left({x}+{y}−\mathrm{1}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{x}\left(−\mathrm{6}+\lambda\right)+{y}\left(−\mathrm{8}+\lambda\right)−\lambda=\mathrm{0} \\ $$$${centre}\:{of}\:{required}\:{cirle}\:{is} \\ $$$${comparing}\:{with}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{gx}+\mathrm{2}{fy}+{c}=\mathrm{0} \\ $$$$\mathrm{2}{gx}=\left(−\mathrm{6}+\lambda\right){x}\rightarrow\mathrm{2}{g}=−\mathrm{6}+\lambda\:\:\:\:\:\:\:{g}=\left(\frac{\lambda−\mathrm{6}}{\mathrm{2}}\right) \\ $$$$\mathrm{2}{fy}=\left(−\mathrm{8}+\lambda\right){y}\rightarrow\mathrm{2}{f}=−\mathrm{8}+\lambda\:\:\:\:\:\:\:{f}=\left(\frac{\lambda−\mathrm{8}}{\mathrm{2}}\right) \\ $$$${centre}\:{of}\:{required}\:{circle}\left(−{g},−{f}\right) \\ $$$$\left(\frac{\mathrm{6}−\lambda}{\mathrm{2}},\frac{\mathrm{8}−\lambda}{\mathrm{2}}\right)\:{which}\:{lies}\:{on}\:{x}+{y}=\mathrm{1} \\ $$$${so}\:\frac{\mathrm{6}−\lambda}{\mathrm{2}}+\frac{\mathrm{8}−\lambda}{\mathrm{2}}=\mathrm{1}\rightarrow\mathrm{7}−\lambda=\mathrm{1}\:\:\:{so}\:\lambda=\mathrm{6} \\ $$$${reqired}\:{circle} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{x}\left(−\mathrm{6}+\mathrm{6}\right)+{y}\left(−\mathrm{8}+\mathrm{6}\right)−\mathrm{6}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{y}−\mathrm{6}=\mathrm{0} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *