Question-88306

Question Number 88306 by TawaTawa1 last updated on 09/Apr/20

Commented by mr W last updated on 10/Apr/20

$${please}\:{try}\:{again}.\:{it}'{s}\:{so}\:{obvious},\:{just} \\ $$$${connect}\:{the}\:{centers}\:{of}\:{both}\:{circles}. \\ $$

Commented by mr W last updated on 09/Apr/20

(8−3−r)^2 +(6−3−r)^2 =(3+r)^2 r^2 −22r+25=0 ⇒r=11−4(√6)≈1.202

$$\left(\mathrm{8}−\mathrm{3}−{r}\right)^{\mathrm{2}} +\left(\mathrm{6}−\mathrm{3}−{r}\right)^{\mathrm{2}} =\left(\mathrm{3}+{r}\right)^{\mathrm{2}} \\ $$$${r}^{\mathrm{2}} −\mathrm{22}{r}+\mathrm{25}=\mathrm{0} \\ $$$$\Rightarrow{r}=\mathrm{11}−\mathrm{4}\sqrt{\mathrm{6}}\approx\mathrm{1}.\mathrm{202} \\ $$

Commented by TawaTawa1 last updated on 09/Apr/20

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

Commented by TawaTawa1 last updated on 09/Apr/20

My problem is how you get the equation sir. I can expand and simplify.

$$\mathrm{My}\:\mathrm{problem}\:\mathrm{is}\:\mathrm{how}\:\mathrm{you}\:\mathrm{get}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{can}\:\mathrm{expand}\:\mathrm{and}\:\mathrm{simplify}. \\ $$

Commented by TawaTawa1 last updated on 10/Apr/20

$$\mathrm{Alright}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{will}\:\mathrm{do}\:\mathrm{that} \\ $$

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Question-88306

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