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Question Number 215141 by cherokeesay last updated on 29/Dec/24
Answered by aleks041103 last updated on 30/Dec/24
eqn: (x−s)^2 +y^2 =r^2 the following systems must have one soln { ((y=(√x))),(((x−s)^2 +y^2 =r^2 )) :} and { ((4x+4y=17)),(((x−s)^2 +y^2 =r^2 )) :} also 0<s<17/4 1st: x^2 +(1−2s)x+s^2 −r^2 =0 ⇒one soln if (1−2s)^2 −4(s^2 −r^2 )=0 ⇒1+4s^2 −4s−4s^2 +4r^2 =0 ⇒4r^2 −4s+1=0 ⇒s−(1/4)=r^2 ⇒eqn: (x−s)^2 +y^2 =s−(1/4) 4x+4y=17 is tangent to the circle ⇒the tangent is st 45° ⇒s+(√2)r=((17)/4) ⇒(1/4)+r^2 +(√2)r=((17)/4)⇒r^2 +(√2)r−4=0 r_(1/2) =((−(√2)±(√(2+16)))/2)=((√2)/2)(−1±3) r>0⇒r=(√2) ⇒S=(1/2)πr^2 =π ⇒S=π, (x−(9/4))^2 +y^2 =2 Alternayively: (x−(1/4)−(s−1/4))^2 +y^2 =(s−1/4) ⇒4(x−1/4)+4y=16 ⇒(x−(1/4))+y=4 z=x−1/4 ⇒(z−r^2 )^2 +y^2 =r^2 and z+y=4 ⇒(z−r^2 )^2 +(4−z)^2 =r^2 ⇒z^2 −2zr^2 −r^2 +r^4 +16−8z+z^2 =0 ⇒2z^2 −2(r^2 +4)z+16−r^2 +r^4 =0 one soln for z iff (r^2 +4)^2 +2(−16−r^4 +r^2 )=0 ⇒r^4 +8r^2 +16−32−2r^4 +2r^2 =0 ⇒−r^4 +10r^2 −16=0⇒r^4 −10r^2 +16=0 ⇒r^2 =((10±(√(10^2 −4.16)))/2)=((10±6)/2)=5±3=2,8 x<((17)/4)⇒z<4 ⇒ r^2 =2
$${eqn}:\:\left({x}−{s}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${the}\:{following}\:{systems}\:{must}\:{have}\:{one}\:{soln} \\ $$$$\begin{cases}{{y}=\sqrt{{x}}}\\{\left({x}−{s}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={r}^{\mathrm{2}} }\end{cases}\:\:\:{and}\:\:\:\begin{cases}{\mathrm{4}{x}+\mathrm{4}{y}=\mathrm{17}}\\{\left({x}−{s}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={r}^{\mathrm{2}} }\end{cases} \\ $$$${also}\:\mathrm{0}<{s}<\mathrm{17}/\mathrm{4} \\ $$$$\mathrm{1}{st}:\:{x}^{\mathrm{2}} +\left(\mathrm{1}−\mathrm{2}{s}\right){x}+{s}^{\mathrm{2}} −{r}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{one}\:{soln}\:{if}\:\left(\mathrm{1}−\mathrm{2}{s}\right)^{\mathrm{2}} −\mathrm{4}\left({s}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{1}+\mathrm{4}{s}^{\mathrm{2}} −\mathrm{4}{s}−\mathrm{4}{s}^{\mathrm{2}} +\mathrm{4}{r}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\mathrm{4}{r}^{\mathrm{2}} −\mathrm{4}{s}+\mathrm{1}=\mathrm{0}\:\Rightarrow{s}−\frac{\mathrm{1}}{\mathrm{4}}={r}^{\mathrm{2}} \\ $$$$\Rightarrow{eqn}:\:\left({x}−{s}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={s}−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{4}{x}+\mathrm{4}{y}=\mathrm{17}\:{is}\:{tangent}\:{to}\:{the}\:{circle} \\ $$$$\Rightarrow{the}\:{tangent}\:{is}\:{st}\:\mathrm{45}° \\ $$$$\Rightarrow{s}+\sqrt{\mathrm{2}}{r}=\frac{\mathrm{17}}{\mathrm{4}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{4}}+{r}^{\mathrm{2}} +\sqrt{\mathrm{2}}{r}=\frac{\mathrm{17}}{\mathrm{4}}\Rightarrow{r}^{\mathrm{2}} +\sqrt{\mathrm{2}}{r}−\mathrm{4}=\mathrm{0} \\ $$$${r}_{\mathrm{1}/\mathrm{2}} =\frac{−\sqrt{\mathrm{2}}\pm\sqrt{\mathrm{2}+\mathrm{16}}}{\mathrm{2}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(−\mathrm{1}\pm\mathrm{3}\right) \\ $$$${r}>\mathrm{0}\Rightarrow{r}=\sqrt{\mathrm{2}} \\ $$$$\Rightarrow{S}=\frac{\mathrm{1}}{\mathrm{2}}\pi{r}^{\mathrm{2}} =\pi \\ $$$$\Rightarrow{S}=\pi,\:\left({x}−\frac{\mathrm{9}}{\mathrm{4}}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{2} \\ $$$$ \\ $$$${Alternayively}: \\ $$$$\left({x}−\frac{\mathrm{1}}{\mathrm{4}}−\left({s}−\mathrm{1}/\mathrm{4}\right)\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\left({s}−\mathrm{1}/\mathrm{4}\right) \\ $$$$\Rightarrow\mathrm{4}\left({x}−\mathrm{1}/\mathrm{4}\right)+\mathrm{4}{y}=\mathrm{16} \\ $$$$\Rightarrow\left({x}−\frac{\mathrm{1}}{\mathrm{4}}\right)+{y}=\mathrm{4} \\ $$$${z}={x}−\mathrm{1}/\mathrm{4} \\ $$$$\Rightarrow\left({z}−{r}^{\mathrm{2}} \right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={r}^{\mathrm{2}} \:{and}\:{z}+{y}=\mathrm{4} \\ $$$$\Rightarrow\left({z}−{r}^{\mathrm{2}} \right)^{\mathrm{2}} +\left(\mathrm{4}−{z}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\Rightarrow{z}^{\mathrm{2}} −\mathrm{2}{zr}^{\mathrm{2}} −{r}^{\mathrm{2}} +{r}^{\mathrm{4}} +\mathrm{16}−\mathrm{8}{z}+{z}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}{z}^{\mathrm{2}} −\mathrm{2}\left({r}^{\mathrm{2}} +\mathrm{4}\right){z}+\mathrm{16}−{r}^{\mathrm{2}} +{r}^{\mathrm{4}} =\mathrm{0} \\ $$$${one}\:{soln}\:{for}\:{z}\:{iff} \\ $$$$\left({r}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} +\mathrm{2}\left(−\mathrm{16}−{r}^{\mathrm{4}} +{r}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow{r}^{\mathrm{4}} +\mathrm{8}{r}^{\mathrm{2}} +\mathrm{16}−\mathrm{32}−\mathrm{2}{r}^{\mathrm{4}} +\mathrm{2}{r}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow−{r}^{\mathrm{4}} +\mathrm{10}{r}^{\mathrm{2}} −\mathrm{16}=\mathrm{0}\Rightarrow{r}^{\mathrm{4}} −\mathrm{10}{r}^{\mathrm{2}} +\mathrm{16}=\mathrm{0} \\ $$$$\Rightarrow{r}^{\mathrm{2}} =\frac{\mathrm{10}\pm\sqrt{\mathrm{10}^{\mathrm{2}} −\mathrm{4}.\mathrm{16}}}{\mathrm{2}}=\frac{\mathrm{10}\pm\mathrm{6}}{\mathrm{2}}=\mathrm{5}\pm\mathrm{3}=\mathrm{2},\mathrm{8} \\ $$$${x}<\frac{\mathrm{17}}{\mathrm{4}}\Rightarrow{z}<\mathrm{4}\:\Rightarrow\:{r}^{\mathrm{2}} =\mathrm{2} \\ $$
Commented by aleks041103 last updated on 30/Dec/24
Commented by cherokeesay last updated on 30/Dec/24
so nice ! thank you master !
$${so}\:{nice}\:! \\ $$$${thank}\:{you}\:{master}\:! \\ $$
Answered by mr W last updated on 30/Dec/24
Commented by mr W last updated on 30/Dec/24
in the rotated coordinate system: y=x^2 say P(p, p^2 ) tan θ=(dy/dx)=2p x_C =p−r sin θ=0 ⇒p=r sin θ=((2pr)/( (√(1+4p^2 )))) ⇒1=((2r)/( (√(1+4p^2 )))) ⇒p=(√(r^2 −(1/4))) y_C =p^2 +r cos θ=p^2 +(r/( (√(1+4p^2 )))) =r^2 −(1/4)+(1/2)=r^2 +(1/4) x_Q =−(r/( (√2))) y_Q =r^2 +(1/4)+(r/( (√2))) −4(−(r/( (√2))))+4(r^2 +(1/4)+(r/( (√2))))=17 r^2 +(√2)r−4=0 r=((−(√2)+3(√2))/2)=(√2) ✓ area of semi−circle =((πr^2 )/2)=π ✓
$${in}\:{the}\:{rotated}\:{coordinate}\:{system}: \\ $$$${y}={x}^{\mathrm{2}} \\ $$$${say}\:{P}\left({p},\:{p}^{\mathrm{2}} \right) \\ $$$$\mathrm{tan}\:\theta=\frac{{dy}}{{dx}}=\mathrm{2}{p} \\ $$$${x}_{{C}} ={p}−{r}\:\mathrm{sin}\:\theta=\mathrm{0}\: \\ $$$$\Rightarrow{p}={r}\:\mathrm{sin}\:\theta=\frac{\mathrm{2}{pr}}{\:\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }} \\ $$$$\Rightarrow\mathrm{1}=\frac{\mathrm{2}{r}}{\:\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }}\:\:\Rightarrow{p}=\sqrt{{r}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$${y}_{{C}} ={p}^{\mathrm{2}} +{r}\:\mathrm{cos}\:\theta={p}^{\mathrm{2}} +\frac{{r}}{\:\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }} \\ $$$$\:\:\:\:\:={r}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}={r}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${x}_{{Q}} =−\frac{{r}}{\:\sqrt{\mathrm{2}}} \\ $$$${y}_{{Q}} ={r}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}+\frac{{r}}{\:\sqrt{\mathrm{2}}} \\ $$$$−\mathrm{4}\left(−\frac{{r}}{\:\sqrt{\mathrm{2}}}\right)+\mathrm{4}\left({r}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}+\frac{{r}}{\:\sqrt{\mathrm{2}}}\right)=\mathrm{17} \\ $$$${r}^{\mathrm{2}} +\sqrt{\mathrm{2}}{r}−\mathrm{4}=\mathrm{0} \\ $$$${r}=\frac{−\sqrt{\mathrm{2}}+\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{2}}=\sqrt{\mathrm{2}}\:\checkmark \\ $$$${area}\:{of}\:{semi}−{circle}\:=\frac{\pi{r}^{\mathrm{2}} }{\mathrm{2}}=\pi\:\checkmark \\ $$
Commented by cherokeesay last updated on 30/Dec/24
great job ! thank so much sir !
$${great}\:{job}\:! \\ $$$${thank}\:{so}\:{much}\:{sir}\:! \\ $$
Commented by mr W last updated on 31/Dec/24
Commented by mr W last updated on 31/Dec/24
y=(√x) ⇒x=y^2 P(p^2 ,p) tan θ=(dx/dy)=2p p=r sin θ=((2pr)/( (√(1+4p^2 )))) ⇒p^2 =r^2 −(1/4) x_C =p^2 +r cos θ=p^2 +(r/( (√(1+4p^2 ))))=r^2 −(1/4)+(1/2)=r^2 +(1/4) x_C =((17)/4)−(√2)r r^2 +(1/4)=((17)/4)−(√2)r r^2 +(√2)r−4=0 ⇒r=(√2)
$${y}=\sqrt{{x}}\:\Rightarrow{x}={y}^{\mathrm{2}} \\ $$$${P}\left({p}^{\mathrm{2}} ,{p}\right) \\ $$$$\mathrm{tan}\:\theta=\frac{{dx}}{{dy}}=\mathrm{2}{p} \\ $$$${p}={r}\:\mathrm{sin}\:\theta=\frac{\mathrm{2}{pr}}{\:\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }}\:\Rightarrow{p}^{\mathrm{2}} ={r}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${x}_{{C}} ={p}^{\mathrm{2}} +{r}\:\mathrm{cos}\:\theta={p}^{\mathrm{2}} +\frac{{r}}{\:\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }}={r}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}={r}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${x}_{{C}} =\frac{\mathrm{17}}{\mathrm{4}}−\sqrt{\mathrm{2}}{r} \\ $$$${r}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{17}}{\mathrm{4}}−\sqrt{\mathrm{2}}{r} \\ $$$${r}^{\mathrm{2}} +\sqrt{\mathrm{2}}{r}−\mathrm{4}=\mathrm{0} \\ $$$$\Rightarrow{r}=\sqrt{\mathrm{2}} \\ $$

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