Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 11221 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 17/Mar/17

Commented by mrW1 last updated on 18/Mar/17

center point of the circle M(a,a) radius of the circle R=(a−1)(√2) sin (θ/2)=(R/(a(√2)))=∣((a−1)/a)∣≤1 θ=2sin^(−1) (∣((a−1)/a)∣), ∣a∣≥(1/2) when ∣a∣→(1/2), θ→180° when ∣a∣→1, θ→0° when ∣a∣→∞, θ→180°

centerpointofthecircleM(a,a)radiusofthecircleR=(a−1)2sinθ2=Ra2=∣a−1a∣⩽1θ=2sin−1(∣a−1a∣),∣a∣⩾12when∣a∣→12,θ→180°when∣a∣→1,θ→0°when∣a∣→∞,θ→180°

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 17/Mar/17

hello mrW1.thank you so much.but i think,this is a very spicial case of quistion.centre of circle my not alwyes lie on the line y=x.

hellomrW1.thankyousomuch.butithink,thisisaveryspicialcaseofquistion.centreofcirclemynotalwyeslieontheliney=x.

Commented by mrW1 last updated on 17/Mar/17

the curve xy=1 is symmetric about the line y=x. for a circle which tangents the curve xy=1 at (1,1) its center must be on the line y=x.

thecurvexy=1issymmetricabouttheliney=x.foracirclewhichtangentsthecurvexy=1at(1,1)itscentermustbeontheliney=x.

Commented by mrW1 last updated on 18/Mar/17

Terms of Service

Privacy Policy

Contact: info@tinkutara.com