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Question Number 100817 by ajfour last updated on 28/Jun/20

Commented by ajfour last updated on 28/Jun/20

$$Findsidesofsquaregivena,b,r.$$

Answered by mr W last updated on 28/Jun/20

$$rightvertex(k,h)$$$$eqn.tangentlinetoellipse:$$$$x-y-k+h=0$$$$eqn.tangentlinetocircle:$$$$x+y-k-h=0$$$$eqn.ofcircle:$$$$x^2+{(y-b-r)}^2=r^2$$$$$$$${(-k+h)}^2=a^2+b^2\Rightarrow k-h=\sqrt{a^2+b^2}$$$${(b+r-k-h)}^2=2r^2\Rightarrow k+h=b+(\sqrt{2}+1)r$$$$\Rightarrow k=\frac{\sqrt{a^2+b^2}+b+(\sqrt{2}+1)r}{2}$$$$\Rightarrow h=\frac{b+(\sqrt{2}+1)r-\sqrt{a^2+b^2}}{2}$$$$\Rightarrow s=\sqrt{2}k=\frac{\sqrt{a^2+b^2}+b+(\sqrt{2}+1)r}{\sqrt{2}}$$

Commented by mr W last updated on 28/Jun/20

Commented by ajfour last updated on 28/Jun/20

$${(-k+h)}^2=a^2+b^2\Rightarrow k-h=\sqrt{a^2+b^2}$$$$someexplanationforthisline$$$$Sir....$$

Commented by mr W last updated on 28/Jun/20

$$ify=mx+ctangentstheellipse,then$$$$a^2{m}^2+b^2=c^2.$$$$ithinkonceyouhaveprovedthis.$$

Commented by ajfour last updated on 29/Jun/20

$$UnderstoodSir,niceandcool!$$

Answered by ajfour last updated on 29/Jun/20

$${\{(b+r+r\sqrt{2})-s\sqrt{2}\}}^2=a^2+b^2$$$$\Rightarrow s=\frac{\sqrt{a^2+b^2}+b+r(1+\sqrt{2})}{\sqrt{2}}.$$

Commented by ajfour last updated on 29/Jun/20

$$yourideaimplementedstraightway,$$$$Sir.$$

Commented by mr W last updated on 29/Jun/20

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