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Question Number 38613 by rish@bh last updated on 27/Jun/18

$$\mathrm{The}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{largest}\mathrm{circle}\mathrm{which}$$$$\mathrm{passes}\mathrm{through}(1,2)\mathrm{and}(3,4)\mathrm{and}\mathrm{lies}$$$$\mathrm{completely}\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{quadrant}\mathrm{is}$$$$\mathrm{A})3$$$$\mathrm{B})2$$$$\mathrm{C})\sqrt{6}$$$$\mathrm{D})2\sqrt{5}$$$$\mathrm{I}\mathrm{got}\mathrm{the}\mathrm{answer}\mathrm{as}2\mathrm{but}\mathrm{the}\mathrm{answer}$$$$\mathrm{given}\mathrm{is}2\sqrt{5}.$$

Answered by ajfour last updated on 27/Jun/18

$$Thelarger{circle}^{\prime }smajorsegment$$$$liesbelowthelinejoiningthe$$$$twopoints.Thecirclethentouches$$$$thex-axis.Thereforelet$$$$theeq.of\mathrm{\perp }bisectorofthe$$$$chordfrom(1,2)to(3,4)is$$$$y-3=2-x\Rightarrow x+y=5$$$$letcenterofcirclebe(h,r)$$$$whererisitsradius.$$$$centreliesonx+y=5$$$$\Rightarrow h+r=5$$$$Further$$$${(h-1)}^2+{(r-2)}^2=r^2$$$$\Rightarrow {(4-r)}^2+{(r-2)}^2=r^2$$$$or{r}^2-12r+20=0$$$${(r-6)}^2=16\Rightarrow r=6\pm 4$$$$r=10,2$$$$ash>0\Rightarrow r=2.$$

Commented by rish@bh last updated on 27/Jun/18

$$\mathrm{Thank}\mathrm{you}$$

Commented by prakash jain last updated on 28/Jun/18

$$ifcircletouchesy-axis$$$${(h-1)}^2+{(r-2)}^2=h^2$$$${(h-1)}^2+{(3-h)}^2=h^2$$$$h^2-8h+10=0$$$$h=\frac{8\pm \sqrt{24}}{2}=4\pm 2\sqrt{3}$$$$r=1\mp 2\sqrt{3}$$$$r=1+2\sqrt{3},h=4-2\sqrt{3}$$$$possiblecenterfortouchy-axis$$$$(4+2\sqrt{3},1-2\sqrt{3})$$$$(4-2\sqrt{3},1+2\sqrt{3})$$$$\mathrm{radius}=4-2\sqrt{3}$$

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