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 72606 by TawaTawa last updated on 30/Oct/19

Answered by mr W last updated on 30/Oct/19

Commented by mr W last updated on 30/Oct/19

BC^2 =12^2 +(2r+1)^2 DB=12 CD=(√((r+1)^2 −r^2 ))=(√(2r+1)) 12^2 +(2r+1)^2 =(12+(√(2r+1)))^2 (2r+1)^2 =24(√(2r+1))+2r+1 let u=(√(2r+1)) u^4 =24u+u^2 u^3 −u−24=0 (u−3)(u^2 +3u+8)=0 ⇒u=3 ⇒(√(2r+1))=3 ⇒r=4

$${BC}^{\mathrm{2}} =\mathrm{12}^{\mathrm{2}} +\left(\mathrm{2}{r}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${DB}=\mathrm{12} \\ $$$${CD}=\sqrt{\left({r}+\mathrm{1}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} }=\sqrt{\mathrm{2}{r}+\mathrm{1}} \\ $$$$\mathrm{12}^{\mathrm{2}} +\left(\mathrm{2}{r}+\mathrm{1}\right)^{\mathrm{2}} =\left(\mathrm{12}+\sqrt{\mathrm{2}{r}+\mathrm{1}}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{2}{r}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{24}\sqrt{\mathrm{2}{r}+\mathrm{1}}+\mathrm{2}{r}+\mathrm{1} \\ $$$${let}\:{u}=\sqrt{\mathrm{2}{r}+\mathrm{1}} \\ $$$${u}^{\mathrm{4}} =\mathrm{24}{u}+{u}^{\mathrm{2}} \\ $$$${u}^{\mathrm{3}} −{u}−\mathrm{24}=\mathrm{0} \\ $$$$\left({u}−\mathrm{3}\right)\left({u}^{\mathrm{2}} +\mathrm{3}{u}+\mathrm{8}\right)=\mathrm{0} \\ $$$$\Rightarrow{u}=\mathrm{3} \\ $$$$\Rightarrow\sqrt{\mathrm{2}{r}+\mathrm{1}}=\mathrm{3} \\ $$$$\Rightarrow{r}=\mathrm{4} \\ $$

Commented by TawaTawa last updated on 30/Oct/19

God bless you sir.

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

Answered by Tanmay chaudhury last updated on 30/Oct/19

circle x^2 +(y−r)^2 =r^2 (x/(12))+(y/(2r+1))=1 is tangent→BC distance from(0,r) to BC is radius (2r+1)x+12y−12(2r+1)=0 ∣(((2r+1)×0+12(r)−24r−12)/(√((2r+1)^2 +12^2 )))∣=r −12r−12=r×(√(4r^2 +4r+1+144)) 144r^2 +144+288r=4r^4 +4r^3 +145r^2 4r^4 +4r^3 +r^2 −288r−144=0 r^2 (4r^2 +4r+1)−144(2r+1)=0 r^2 (2r+1)^2 −144(2r+1)=0 (2r+1)(2r^3 +r^2 −144)=0 r≠−0.5 2r^3 +r^2 −128−16=0 2(r^3 −64)+(r+4)(r−4)=0 2(r−4)(r^2 +4r+16)+(r+4)(r−4)=0 (r−4)(2r^2 +8r+32+r+4)=0 so r=4 answer

$${circle}\:{x}^{\mathrm{2}} +\left({y}−{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\frac{{x}}{\mathrm{12}}+\frac{{y}}{\mathrm{2}{r}+\mathrm{1}}=\mathrm{1}\:{is}\:{tangent}\rightarrow{BC} \\ $$$${distance}\:{from}\left(\mathrm{0},{r}\right)\:{to}\:{BC}\:{is}\:{radius} \\ $$$$\left(\mathrm{2}{r}+\mathrm{1}\right){x}+\mathrm{12}{y}−\mathrm{12}\left(\mathrm{2}{r}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\mid\frac{\left(\mathrm{2}{r}+\mathrm{1}\right)×\mathrm{0}+\mathrm{12}\left({r}\right)−\mathrm{24}{r}−\mathrm{12}}{\sqrt{\left(\mathrm{2}{r}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} }}\mid={r} \\ $$$$−\mathrm{12}{r}−\mathrm{12}={r}×\sqrt{\mathrm{4}{r}^{\mathrm{2}} +\mathrm{4}{r}+\mathrm{1}+\mathrm{144}}\: \\ $$$$\mathrm{144}{r}^{\mathrm{2}} +\mathrm{144}+\mathrm{288}{r}=\mathrm{4}{r}^{\mathrm{4}} +\mathrm{4}{r}^{\mathrm{3}} +\mathrm{145}{r}^{\mathrm{2}} \\ $$$$\mathrm{4}{r}^{\mathrm{4}} +\mathrm{4}{r}^{\mathrm{3}} +{r}^{\mathrm{2}} −\mathrm{288}{r}−\mathrm{144}=\mathrm{0} \\ $$$${r}^{\mathrm{2}} \left(\mathrm{4}{r}^{\mathrm{2}} +\mathrm{4}{r}+\mathrm{1}\right)−\mathrm{144}\left(\mathrm{2}{r}+\mathrm{1}\right)=\mathrm{0} \\ $$$${r}^{\mathrm{2}} \left(\mathrm{2}{r}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{144}\left(\mathrm{2}{r}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\left(\mathrm{2}{r}+\mathrm{1}\right)\left(\mathrm{2}{r}^{\mathrm{3}} +{r}^{\mathrm{2}} −\mathrm{144}\right)=\mathrm{0} \\ $$$${r}\neq−\mathrm{0}.\mathrm{5} \\ $$$$\mathrm{2}{r}^{\mathrm{3}} +{r}^{\mathrm{2}} −\mathrm{128}−\mathrm{16}=\mathrm{0} \\ $$$$\mathrm{2}\left({r}^{\mathrm{3}} −\mathrm{64}\right)+\left({r}+\mathrm{4}\right)\left({r}−\mathrm{4}\right)=\mathrm{0} \\ $$$$\mathrm{2}\left({r}−\mathrm{4}\right)\left({r}^{\mathrm{2}} +\mathrm{4}{r}+\mathrm{16}\right)+\left({r}+\mathrm{4}\right)\left({r}−\mathrm{4}\right)=\mathrm{0} \\ $$$$\left({r}−\mathrm{4}\right)\left(\mathrm{2}{r}^{\mathrm{2}} +\mathrm{8}{r}+\mathrm{32}+{r}+\mathrm{4}\right)=\mathrm{0} \\ $$$${so}\:\:{r}=\mathrm{4}\:{answer} \\ $$

Commented by TawaTawa last updated on 30/Oct/19

God bless you sir.

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com