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Question-58879




Question Number 58879 by gunawan last updated on 01/May/19
Answered by mr W last updated on 01/May/19
Commented by mr W last updated on 01/May/19
assume B(b,0)  ⇒D(3, 3+b)  D is on the circle,  ⇒3^2 +(3+b)^2 =5^2   ⇒b^2 +6b−7=0  ⇒(b−1)(b−7)=0  ⇒b=1  C(b+3, b)=(4, 1)  ⇒x=(√((5−4)^2 +1^2 ))=(√2)
$${assume}\:{B}\left({b},\mathrm{0}\right) \\ $$$$\Rightarrow{D}\left(\mathrm{3},\:\mathrm{3}+{b}\right) \\ $$$${D}\:{is}\:{on}\:{the}\:{circle}, \\ $$$$\Rightarrow\mathrm{3}^{\mathrm{2}} +\left(\mathrm{3}+{b}\right)^{\mathrm{2}} =\mathrm{5}^{\mathrm{2}} \\ $$$$\Rightarrow{b}^{\mathrm{2}} +\mathrm{6}{b}−\mathrm{7}=\mathrm{0} \\ $$$$\Rightarrow\left({b}−\mathrm{1}\right)\left({b}−\mathrm{7}\right)=\mathrm{0} \\ $$$$\Rightarrow{b}=\mathrm{1} \\ $$$${C}\left({b}+\mathrm{3},\:{b}\right)=\left(\mathrm{4},\:\mathrm{1}\right) \\ $$$$\Rightarrow{x}=\sqrt{\left(\mathrm{5}−\mathrm{4}\right)^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }=\sqrt{\mathrm{2}} \\ $$

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