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Question Number 45785 by Tawa1 last updated on 16/Oct/18

Answered by tanmay.chaudhury50@gmail.com last updated on 16/Oct/18

let centre of circle is (α,β) and radius r so eqn of circle (x−α)^2 +(y−β)^2 =r^2 now as per condition 1)x axis is[the tangent of circle so distance from cente to[ x sxis is the radius. distance from centre(α,β) to xaxis is β=radius β=r so eqn is (x−α)^2 +(y−β)^2 =β^2 2)centre lies on 2y=x 2β=α 3)(14,2) liez on circle so (14−α)^2 +(2−β)^2 =β^2 (14−2β)^2 +(2−β)^2 =β^2 given α=2β 196−56β+4β^2 +4−4β+β^2 =β^2 4β^2 −60β+200=0 β^2 −15β+50=0 (β−5)(β−10)=0 β=5 α=10 eqn (x−10)^2 +(y−5)^2 =5^2 when β=10 α=20 eqn (x−20)^2 +(y−10)^2 =10^2

$${let}\:{centre}\:{of}\:{circle}\:{is}\:\left(\alpha,\beta\right)\:{and}\:{radius}\:{r} \\ $$$${so}\:{eqn}\:{of}\:{circle}\:\left({x}−\alpha\right)^{\mathrm{2}} +\left({y}−\beta\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${now}\:{as}\:{per}\:{condition} \\ $$$$\left.\mathrm{1}\right){x}\:{axis}\:{is}\left[{the}\:{tangent}\:{of}\:{circle}\:{so}\:{distance}\right. \\ $$$${from}\:{cente}\:{to}\left[\:{x}\:{sxis}\:\:{is}\:{the}\:{radius}.\right. \\ $$$${distance}\:{from}\:{centre}\left(\alpha,\beta\right)\:{to}\:{xaxis}\:{is}\:\beta={radius} \\ $$$$\beta={r}\: \\ $$$${so}\:{eqn}\:{is}\:\left({x}−\alpha\right)^{\mathrm{2}} +\left({y}−\beta\right)^{\mathrm{2}} =\beta^{\mathrm{2}} \\ $$$$\left.\mathrm{2}\right){centre}\:{lies}\:{on}\:\mathrm{2}{y}={x} \\ $$$$\mathrm{2}\beta=\alpha \\ $$$$\left.\mathrm{3}\right)\left(\mathrm{14},\mathrm{2}\right)\:{liez}\:{on}\:{circle}\:{so}\: \\ $$$$\left(\mathrm{14}−\alpha\right)^{\mathrm{2}} +\left(\mathrm{2}−\beta\right)^{\mathrm{2}} =\beta^{\mathrm{2}} \\ $$$$\left(\mathrm{14}−\mathrm{2}\beta\right)^{\mathrm{2}} +\left(\mathrm{2}−\beta\right)^{\mathrm{2}} =\beta^{\mathrm{2}} \:\:\:\:{given}\:\alpha=\mathrm{2}\beta \\ $$$$\mathrm{196}−\mathrm{56}\beta+\mathrm{4}\beta^{\mathrm{2}} +\mathrm{4}−\mathrm{4}\beta+\beta^{\mathrm{2}} =\beta^{\mathrm{2}} \\ $$$$\mathrm{4}\beta^{\mathrm{2}} −\mathrm{60}\beta+\mathrm{200}=\mathrm{0} \\ $$$$\beta^{\mathrm{2}} −\mathrm{15}\beta+\mathrm{50}=\mathrm{0} \\ $$$$\left(\beta−\mathrm{5}\right)\left(\beta−\mathrm{10}\right)=\mathrm{0} \\ $$$$\beta=\mathrm{5}\:\:\alpha=\mathrm{10}\:\:{eqn}\:\left({x}−\mathrm{10}\right)^{\mathrm{2}} +\left({y}−\mathrm{5}\right)^{\mathrm{2}} =\mathrm{5}^{\mathrm{2}} \\ $$$${when}\:\beta=\mathrm{10}\:\alpha=\mathrm{20} \\ $$$${eqn}\:\left({x}−\mathrm{20}\right)^{\mathrm{2}} +\left({y}−\mathrm{10}\right)^{\mathrm{2}} =\mathrm{10}^{\mathrm{2}} \\ $$$$ \\ $$

Commented by Tawa1 last updated on 16/Oct/18

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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