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Question Number 166875 by amin96 last updated on 01/Mar/22

Commented by amin96 last updated on 01/Mar/22

$$\mathbf{circle}\mathbf{radius}=???$$

Answered by MJS_new last updated on 01/Mar/22

$$\mathrm{we}\mathrm{only}\mathrm{need}\mathrm{the}\mathrm{triangle}\mathrm{with}\mathrm{sides}$$$$a=3b=4c=5$$$$\mathrm{let}\mathrm{the}\mathrm{right}\mathrm{angled}\mathrm{vertice}=\mathrm{the}\mathrm{origin}$$$$\mathrm{the}\mathrm{circle}\mathrm{then}\mathrm{has}\mathrm{the}\mathrm{equation}$$$${(x-r)}^2+{(y+r)}^2=r^2$$$$\mathrm{and}\mathrm{the}\mathrm{bevel}\mathrm{line}$$$$y=\frac{4}{3}x-4$$$$\Rightarrow$$$${(x-r)}^2+{(\frac{4}{3}x-4+r)}^2-r^2=0$$$$\iff$$$$x^2+\frac{6(r-16)}{25}x+\frac{9{(r-4)}^2}{25}=0$$$$\mathrm{exactly}1\mathrm{solution}\iff \mathrm{\Delta }=0$$$$r^2-7r+6=0$$$$\Rightarrow r=1\left[\mathrm{the}\mathrm{incircle}\mathrm{of}\mathrm{the}\mathrm{triangle}\right]$$$$r=6$$

Commented by MJS_new last updated on 01/Mar/22

$$\mathrm{you}\mathrm{could}\mathrm{also}\mathrm{use}\mathrm{the}\mathrm{formula}\mathrm{for}\mathrm{the}$$$$\mathrm{excircle}\mathrm{radius}$$$$r_c=\frac{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}{2(a+b-c)}$$

Commented by Tawa11 last updated on 01/Mar/22

$$\mathrm{Great}\mathrm{sir}$$

Answered by mr W last updated on 01/Mar/22

Commented by mr W last updated on 01/Mar/22

$$r-3+r-4=5$$$$\Rightarrow r=6$$

Commented by Tawa11 last updated on 01/Mar/22

$$\mathrm{Great}\mathrm{sir}$$

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