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Question Number 118011 by mmmmmm1 last updated on 14/Oct/20

Answered by mr W last updated on 14/Oct/20

$${(x-h)}^2+{(y-k)}^2=R^2$$$$$$$${(0-h)}^2+{(0-k)}^2=R^2...\left(i\right)$$$${(-4-h)}^2+{(0-k)}^2=R^2...\left(ii\right)$$$${(2-h)}^2+{(-3-k)}^2=R^2...\left(iii\right)$$$$\left(ii\right)-\left(i\right):$$$$4(4+2h)=0\Rightarrow h=-2$$$$\left(iii\right)-\left(i\right):$$$$2(2-2h)+3(3+2k)=0\Rightarrow k=-\frac{7}{2}$$$$R^2={\left(2\right)}^2+{\left(\frac{7}{2}\right)}^2=\frac{65}{4}$$$$\Rightarrow R=\frac{\sqrt{65}}{2}$$

Answered by MJS_new last updated on 14/Oct/20

$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{the}\mathrm{circumcircle}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{with}$$$$a=4$$$$b=\sqrt{2^2+3^2}=\sqrt{13}$$$$c=\sqrt{{(2+4)}^2+3^2}=\sqrt{45}$$$$R=\frac{abc}{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}=\frac{\sqrt{65}}{2}$$$$\Rightarrow \mathrm{area}\mathrm{of}\mathrm{circle}=\frac{65\pi }{4}$$

Answered by mr W last updated on 14/Oct/20

$$anotherway:$$$$a=4$$$$b=\sqrt{2^2+3^2}=\sqrt{13}$$$$c=\sqrt{3^2+{(2+4)}^2}=\sqrt{45}$$$$\mathrm{\Delta }=\frac{4\times 3}{2}=6$$$$R=\frac{abc}{4\mathrm{\Delta }}=\frac{4\times \sqrt{13}\times \sqrt{45}}{4\times 6}=\frac{\sqrt{65}}{2}$$

Answered by 1549442205PVT last updated on 15/Oct/20

Commented by 1549442205PVT last updated on 15/Oct/20

$$\mathrm{Given}\mathrm{AC}=4,\mathrm{CD}=3,\mathrm{DB}=2$$$$\widehat{\mathrm{ACD}}=\widehat{\mathrm{CDB}}=90°$$$$\mathrm{first}\mathrm{way}:$$$$\mathrm{AC}//\mathrm{BD}\Rightarrow \frac{\mathrm{DH}}{\mathrm{HC}}=\frac{\mathrm{BD}}{\mathrm{AC}}=\frac{2}{4}=\frac{1}{2}$$$$\mathrm{CD}=3\left(\mathrm{hypothesis}\right)\Rightarrow \mathrm{HC}=2$$$$\mathrm{AH}=\sqrt{{\mathrm{AC}}^2+{\mathrm{CH}}^2}=\sqrt{4^2+2^2}=2\sqrt{5}$$$$\sin \widehat{\mathrm{CAH}}=\frac{\mathrm{CH}}{\mathrm{AH}}=\frac{2}{2\sqrt{5}}=\frac{1}{\sqrt{5}}$$$$\sin \widehat{\mathrm{CLB}}=\sin \widehat{\mathrm{CAB}}=\sin \widehat{\mathrm{CAH}}=1/\sqrt{5}$$$$\mathrm{\Delta }\mathrm{BCL}\mathrm{is}\mathrm{right}\mathrm{at}\mathrm{B}\Rightarrow 2\mathrm{R}=\mathrm{CL}=\frac{\mathrm{BC}}{\sin \widehat{\mathrm{CLB}}}$$$$=\frac{\sqrt{3^2+2^2}}{\left(1/\sqrt{5}\right)}=\sqrt{65}\Rightarrow \mathrm{R}=\sqrt{65}/2$$$$\Rightarrow {\mathrm{S}}_{\mathrm{circle}}=\pi {\mathrm{R}}^2=\frac{65\pi }{4}$$$$\mathrm{second}\mathrm{way}:\mathrm{Suppose}\left(\mathrm{CD}\right)\cap \left(\mathrm{O}\right)=\mathrm{E}$$$$\mathrm{Denote}\mathrm{by}\mathrm{I},\mathrm{F}\mathrm{midpoint}\mathrm{AC},\mathrm{CE}\mathrm{respectively}$$$$\mathrm{Put}\mathrm{DF}=\mathrm{x}\Rightarrow \mathrm{BF}=\mathrm{x}+3,\mathrm{J}=\left(\mathrm{BD}\right)\cap \mathrm{OI}$$$$\Rightarrow \mathrm{BJ}\mathrm{\perp }\mathrm{OI}(\mathrm{since}\mathrm{OI}\mathrm{\perp }\mathrm{AC},\mathrm{AC}//\mathrm{BJ})$$$$\Rightarrow \mathrm{OJ}=\mathrm{DF}=\mathrm{x},\mathrm{CI}=\mathrm{DJ}=2,\mathrm{BJ}=4$$$$\mathrm{so}\mathrm{in}\mathrm{the}\mathrm{righttriangles}\mathrm{OFC}\mathrm{and}\mathrm{OJB}$$$$\mathrm{we}\mathrm{have}:{(\mathrm{x}+3)}^2+2^2={\mathrm{R}}^2={\mathrm{x}}^2+4^2$$$$\Rightarrow {\mathrm{x}}^2+6\mathrm{x}+13={\mathrm{x}}^2+16\Rightarrow 6\mathrm{x}=3\Rightarrow \mathrm{x}=\frac{1}{2}$$$$\Rightarrow \mathrm{R}=\sqrt{{\mathrm{x}}^2+4^2}=\sqrt{\frac{1}{4}+16}=\sqrt{\frac{65}{4}}=\frac{\sqrt{65}}{2}$$$$\Rightarrow {\mathrm{S}}_{\mathrm{circle}}=\frac{65\pi }{4}$$

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