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Question Number 159392 by ajfour last updated on 16/Nov/21

Commented by ajfour last updated on 16/Nov/21

$$Findradiusofthecircles\left(equal\right).$$

Answered by mr W last updated on 17/Nov/21

Commented by mr W last updated on 18/Nov/21

$$eqn.ofOC:$$$$sayy=mx\Rightarrow mx-y=0$$$$eqn.ofAD:$$$$sauy=-k(x-1)\Rightarrow kx+y-k=0$$$$P(r,1-r)$$$$Q(q,r)$$$$$$$$r=-\frac{mr-1+r}{\sqrt{1+m^2}}...\left(i\right)$$$$r=\frac{mq-r}{\sqrt{1+m^2}}...\left(ii\right)$$$$r=-\frac{kr+1-r-k}{\sqrt{1+k^2}}...\left(iii\right)$$$$r=-\frac{kq+r-k}{\sqrt{1+k^2}}...\left(iv\right)$$$$4unknowns:r,m,k,q$$$$$$$$mq-r+mr-1+r=0$$$$\Rightarrow q=\frac{1}{m}-r$$$$kq+r-k=kr+1-r-k$$$$\Rightarrow k=\frac{1-2r}{q-r}$$$$k=\frac{1-2r}{\frac{1}{m}-2r}...\left(I\right)$$$$$$$$from\left(ii\right):$$$$r=\frac{1-(m+1)r}{\sqrt{1+m^2}}$$$$r=\frac{1}{1+m+\sqrt{1+m^2}}...\left(II\right)$$$$from\left(iv\right):$$$$1-k+\sqrt{1+k^2}=(1-\frac{1}{m})\frac{k}{r}...\left(III\right)$$$$puttingI,IIintoIIIwegetan.eqn.$$$$form.$$$$\Rightarrow m\approx 1.243756$$$$\Rightarrow r\approx 0.2694$$

Commented by mr W last updated on 18/Nov/21

Commented by ajfour last updated on 18/Nov/21

$$Excellentsir,thanks!$$$$imoutofmtowntotakeupajob,$$$$hencemyinconsistency..$$

Commented by mr W last updated on 18/Nov/21

$$goodluckwiththenewjobsir!$$

Commented by Tawa11 last updated on 28/Nov/21

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

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