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The minimum radius of the circle which contains the threecircles, x^(2)+y^(2)-4y-5=0,x^(2)+y^(2)+12x+4y+31=0 and x^(2)+y^(2)+6x+12y+36=0 is |
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Answer» `(7)/(18)sqrt(900)+3` `S_(2)-=x^(2)+y^(2)+12x+4y+31=0` and `S_(3) -=x^(2)+y^(2)+6x+12y+36=0` `C_(1) -= (0,2),r_(1)=3` `C_(2) -= (-6,-2), r_(2)=3` `C_(3) -= (-3,-6) , r_(3)=3` All circles have the same radius, THERADIUS of the CIRCLE touching all the circles is `CP=C C_(1)=C_(1)P+3` Let `C(h,k)` be the center of the required circle. Then, `C C_(1)= C C_(2) = C C_(3)`or `C C_(1)^(2) = C C_(2)^(2)= C C_(3)^(2)` or `(h-0)^(2)+(k-2)^(2)=(h+6)^(2)+(k+2)^(2)` `=(h+3)^(2)+(k+6)^(2)` or `-4k+4=12h+4k+40=6h+12k+45` or `3h+2k+9=0` and `6h-8k-5=0` Solving, we get `h=(-31)/(18),k=(-23)/(12)` `:. C C_(1)=sqrt((0+(31)/(18))^(2)+(2+(23)/(12))^(2))=(5)/(36)sqrt(949)` Now, `CP =C C_(1)+3=(5)/(36)sqrt(949)+3` |
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