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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

letcentreofcircleis(α,β)andradiusrsoeqnofcircle(x−α)2+(y−β)2=r2nowaspercondition1)xaxisis[thetangentofcirclesodistancefromcenteto[xsxisistheradius.distancefromcentre(α,β)toxaxisisβ=radiusβ=rsoeqnis(x−α)2+(y−β)2=β22)centrelieson2y=x2β=α3)(14,2)liezoncircleso(14−α)2+(2−β)2=β2(14−2β)2+(2−β)2=β2givenα=2β196−56β+4β2+4−4β+β2=β24β2−60β+200=0β2−15β+50=0(β−5)(β−10)=0β=5α=10eqn(x−10)2+(y−5)2=52whenβ=10α=20eqn(x−20)2+(y−10)2=102

Commented by Tawa1 last updated on 16/Oct/18

God bless you sir

Godblessyousir

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