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Question Number 174283 by ajfour last updated on 28/Jul/22

Commented by ajfour last updated on 28/Jul/22

Find PQ in terms of a,b,θ.

$${Find}\:{PQ}\:{in}\:{terms}\:{of}\:{a},{b},\theta. \\ $$

Answered by mr W last updated on 29/Jul/22

A=center of circle a B=center of circle b PQ=2r AB=((b−a)/(sin (θ/2)))=d d=(√(a^2 −r^2 ))+(√(b^2 −r^2 )) d^2 +a^2 −r^2 −2d(√(a^2 −r^2 ))=b^2 −r^2 d^2 +a^2 −b^2 =2d(√(a^2 −r^2 )) (d^2 +a^2 −b^2 )^2 =4d^2 (a^2 −r^2 ) r=(√(a^2 −(((d^2 +a^2 −b^2 )/(2d)))^2 )) ⇒PQ=2(√(a^2 −(((d^2 +a^2 −b^2 )/(2d)))^2 ))

$${A}={center}\:{of}\:{circle}\:{a} \\ $$$${B}={center}\:{of}\:{circle}\:{b} \\ $$$${PQ}=\mathrm{2}{r} \\ $$$${AB}=\frac{{b}−{a}}{\mathrm{sin}\:\frac{\theta}{\mathrm{2}}}={d} \\ $$$${d}=\sqrt{{a}^{\mathrm{2}} −{r}^{\mathrm{2}} }+\sqrt{{b}^{\mathrm{2}} −{r}^{\mathrm{2}} } \\ $$$${d}^{\mathrm{2}} +{a}^{\mathrm{2}} −{r}^{\mathrm{2}} −\mathrm{2}{d}\sqrt{{a}^{\mathrm{2}} −{r}^{\mathrm{2}} }={b}^{\mathrm{2}} −{r}^{\mathrm{2}} \\ $$$${d}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\mathrm{2}{d}\sqrt{{a}^{\mathrm{2}} −{r}^{\mathrm{2}} } \\ $$$$\left({d}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{4}{d}^{\mathrm{2}} \left({a}^{\mathrm{2}} −{r}^{\mathrm{2}} \right) \\ $$$${r}=\sqrt{{a}^{\mathrm{2}} −\left(\frac{{d}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{d}}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow{PQ}=\mathrm{2}\sqrt{{a}^{\mathrm{2}} −\left(\frac{{d}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{d}}\right)^{\mathrm{2}} } \\ $$

Commented by ajfour last updated on 29/Jul/22

wonderful, thanks Sir!

$${wonderful},\:{thanks}\:{Sir}! \\ $$

Commented by Tawa11 last updated on 29/Jul/22

Great sir

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

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