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Question Number 58282 by behi83417@gmail.com last updated on 20/Apr/19

$$\mathbf{A}\mathbf{circle}\mathbf{tangents}\mathbf{to}:\mathbf{x}\mathbf{and}\mathbf{y}\mathbf{axes}\mathbf{and}$$$${\mathbf{x}}^{\frac{1}{2}}+{\mathbf{y}}^{\frac{1}{2}}={\mathbf{a}}^{\frac{1}{2}}.\mathbf{find}\mathbf{its}\mathbf{radious}.$$

Answered by mr W last updated on 21/Apr/19

$$duetosymmetrythecircletouches$$$$thecurveat(t,t).$$$$\sqrt{t}+\sqrt{t}=\sqrt{a}$$$$\Rightarrow t=\frac{a}{4}$$$$eqn.ofcircle:{(x-r)}^2+{(y-r)}^2=r^2$$$${(t-r)}^2+{(t-r)}^2=r^2$$$$\sqrt{2}(t-r)=\pm r$$$$\Rightarrow r=\frac{\sqrt{2}}{\sqrt{2}+1}t=(2-\sqrt{2})t=\frac{2-\sqrt{2}}{4}a$$$$or$$$$\Rightarrow r=\frac{\sqrt{2}}{\sqrt{2}-1}t=(2+\sqrt{2})t=\frac{2+\sqrt{2}}{4}a$$$$(butthiscircleintersectsalsowith$$$$thecurve.)$$

Commented by behi83417@gmail.com last updated on 21/Apr/19

$$thanksalotdearmaster.$$$$ihavesomeproblemwithlastline!$$

Commented by mr W last updated on 21/Apr/19

$$thecirclewithr=\frac{2-\sqrt{2}}{4}aonlytangents$$$$thecurve,butthecirclewithr=\frac{2+\sqrt{2}}{4}a$$$$notonlytangentsthecurvebutalso$$$$intersectswithit.$$

Commented by mr W last updated on 21/Apr/19

Commented by behi83417@gmail.com last updated on 21/Apr/19

$$thankyousomuchmaster.$$$$perfect!nowitiscleartome.$$

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