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Question Number 57522 by rahul 19 last updated on 07/Apr/19

The radius of circle having minimum area,which touches the curve y=4−x^2 and the lines y=∣x∣ is ?

$${The}\:{radius}\:{of}\:{circle}\:{having}\:{minimum} \\ $$$${area},{which}\:{touches}\:{the}\:{curve}\:{y}=\mathrm{4}−{x}^{\mathrm{2}} \\ $$$${and}\:{the}\:{lines}\:{y}=\mid{x}\mid\:{is}\:? \\ $$

Commented by rahul 19 last updated on 06/Apr/19

Ans:(((√(34))−2)/(2(√2))).

$${Ans}:\frac{\sqrt{\mathrm{34}}−\mathrm{2}}{\mathrm{2}\sqrt{\mathrm{2}}}. \\ $$

Commented by MJS last updated on 06/Apr/19

Commented by MJS last updated on 06/Apr/19

there are 5 possible circles

$$\mathrm{there}\:\mathrm{are}\:\mathrm{5}\:\mathrm{possible}\:\mathrm{circles} \\ $$

Answered by mr W last updated on 07/Apr/19

for max. circle: (√2)R=4+R ⇒R=(4/((√2)−1))=4((√2)+1)

$${for}\:{max}.\:{circle}: \\ $$$$\sqrt{\mathrm{2}}{R}=\mathrm{4}+{R} \\ $$$$\Rightarrow{R}=\frac{\mathrm{4}}{\sqrt{\mathrm{2}}−\mathrm{1}}=\mathrm{4}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right) \\ $$

Commented by rahul 19 last updated on 07/Apr/19

I′ve edit the Ques. my apologies...

$${I}'{ve}\:{edit}\:{the}\:{Ques}. \\ $$$${my}\:{apologies}... \\ $$

Commented by mr W last updated on 07/Apr/19

Answered by mr W last updated on 07/Apr/19

for min. circle: eqn. of circle x^2 +(y−(√2)r)^2 =r^2 since y=4−x^2 ⇒4−y+y^2 −2(√2)ry+2r^2 =r^2 ⇒y^2 −(2(√2)r+1)y+r^2 +4=0 Δ=(2(√2)r+1)^2 −4(r^2 +4)=0 8r^2 +4(√2)r+1−4r^2 −16=0 4r^2 +4(√2)r−15=0 ⇒r=(((√(17))−(√2))/2)

$${for}\:{min}.\:{circle}: \\ $$$${eqn}.\:{of}\:{circle}\:{x}^{\mathrm{2}} +\left({y}−\sqrt{\mathrm{2}}{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${since}\:{y}=\mathrm{4}−{x}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{4}−{y}+{y}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{2}}{ry}+\mathrm{2}{r}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\Rightarrow{y}^{\mathrm{2}} −\left(\mathrm{2}\sqrt{\mathrm{2}}{r}+\mathrm{1}\right){y}+{r}^{\mathrm{2}} +\mathrm{4}=\mathrm{0} \\ $$$$\Delta=\left(\mathrm{2}\sqrt{\mathrm{2}}{r}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}\left({r}^{\mathrm{2}} +\mathrm{4}\right)=\mathrm{0} \\ $$$$\mathrm{8}{r}^{\mathrm{2}} +\mathrm{4}\sqrt{\mathrm{2}}{r}+\mathrm{1}−\mathrm{4}{r}^{\mathrm{2}} −\mathrm{16}=\mathrm{0} \\ $$$$\mathrm{4}{r}^{\mathrm{2}} +\mathrm{4}\sqrt{\mathrm{2}}{r}−\mathrm{15}=\mathrm{0} \\ $$$$\Rightarrow{r}=\frac{\sqrt{\mathrm{17}}−\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$

Commented by MJS last updated on 07/Apr/19

(((√(34))−2)/(2(√2)))=(((√2)×(√(2×17))−2(√2))/4)=((2(√(17))−2(√2))/4)= =(((√(17))−(√2))/2)

$$\frac{\sqrt{\mathrm{34}}−\mathrm{2}}{\mathrm{2}\sqrt{\mathrm{2}}}=\frac{\sqrt{\mathrm{2}}×\sqrt{\mathrm{2}×\mathrm{17}}−\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{4}}=\frac{\mathrm{2}\sqrt{\mathrm{17}}−\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{4}}= \\ $$$$=\frac{\sqrt{\mathrm{17}}−\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$

Commented by mr W last updated on 07/Apr/19

Commented by rahul 19 last updated on 07/Apr/19

Thank you sir!

$${Thank}\:{you}\:{sir}! \\ $$

Answered by mr W last updated on 07/Apr/19

third possibility: eqn. of circle x^2 +(y+r)^2 =r^2 4−y+y^2 +2ry+r^2 =r^2 y^2 +(2r−1)y+4= Δ=(2r−1)^2 −4×4=0 2r−1=4 ⇒r=(5/2)

$${third}\:{possibility}: \\ $$$${eqn}.\:{of}\:{circle}\:{x}^{\mathrm{2}} +\left({y}+{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\mathrm{4}−{y}+{y}^{\mathrm{2}} +\mathrm{2}{ry}+{r}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${y}^{\mathrm{2}} +\left(\mathrm{2}{r}−\mathrm{1}\right){y}+\mathrm{4}= \\ $$$$\Delta=\left(\mathrm{2}{r}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}×\mathrm{4}=\mathrm{0} \\ $$$$\mathrm{2}{r}−\mathrm{1}=\mathrm{4} \\ $$$$\Rightarrow{r}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$

Commented by mr W last updated on 07/Apr/19

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