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

$$Theradiusofcirclehavingminimum$$$$area,whichtouchesthecurvey=4-x^2$$$$andthelinesy=\mid x\mid is?$$

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

$$Ans:\frac{\sqrt{34}-2}{2\sqrt{2}}.$$

Commented by MJS last updated on 06/Apr/19

Commented by MJS last updated on 06/Apr/19

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

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

$$formax.circle:$$$$\sqrt{2}R=4+R$$$$\Rightarrow R=\frac{4}{\sqrt{2}-1}=4(\sqrt{2}+1)$$

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

$$I^{\prime }veedittheQues.$$$$myapologies...$$

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

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

$$formin.circle:$$$$eqn.ofcircle{x}^2+{(y-\sqrt{2}r)}^2=r^2$$$$sincey=4-x^2$$$$\Rightarrow 4-y+y^2-2\sqrt{2}ry+2r^2=r^2$$$$\Rightarrow y^2-(2\sqrt{2}r+1)y+r^2+4=0$$$$\mathrm{\Delta }={(2\sqrt{2}r+1)}^2-4(r^2+4)=0$$$$8r^2+4\sqrt{2}r+1-4r^2-16=0$$$$4r^2+4\sqrt{2}r-15=0$$$$\Rightarrow r=\frac{\sqrt{17}-\sqrt{2}}{2}$$

Commented by MJS last updated on 07/Apr/19

$$\frac{\sqrt{34}-2}{2\sqrt{2}}=\frac{\sqrt{2}\times \sqrt{2\times 17}-2\sqrt{2}}{4}=\frac{2\sqrt{17}-2\sqrt{2}}{4}=$$$$=\frac{\sqrt{17}-\sqrt{2}}{2}$$

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

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

$$Thankyousir!$$

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

$$thirdpossibility:$$$$eqn.ofcircle{x}^2+{(y+r)}^2=r^2$$$$4-y+y^2+2ry+r^2=r^2$$$$y^2+(2r-1)y+4=$$$$\mathrm{\Delta }={(2r-1)}^2-4\times 4=0$$$$2r-1=4$$$$\Rightarrow r=\frac{5}{2}$$

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

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