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Let each of the circles `S_(1)-=x^(2)+y^(2)+4y-1=0` `S_(1)-= x^(2)+y^(2)+6x+y+8=0` `S_(3)-=x^(2)+y^(2)-4x-4y-37=0` touch the other two. Also, let `P_(1),P_(2)` and `P_(3)` be the points of contact of `S_(1)` and `S_(2) , S_(2)` and `S_(3)`, and `S_(3)` , respectively, `C_(1),C_(2)` and `C_(3)` are the centres of `S_(1),S_(2)` and `S_(3)` respectively. `P_(2)` and `P_(3)` are images of each other with respect to the lineA. `y=x`B. `y = -x`C. `y=x+1`D. `y= -x+2` |
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Answer» Correct Answer - 1 `S_(1)-= x^(2)+y^(2)+4y-1=0` `S_(2) -= x^(2)+y^(2)+6x+y+8=0` `S_(3)-=x^(2)+y^(2)-4x-4y-37=0` `C_(1) -= (0,-2),r_(1)=sqrt(5)` `C_(2) -= (-3,(-1)/(2)),r_(2)=(sqrt(5))/(2)` `C_(3)-= (2,2), r_(3)= 3 sqrt(5)` Also, `C_(1)C_(2)=sqrt(9+(9)/(4))=(3sqrt(5))/(2)=r_(1)+r_(2)` So, `S_(1)` and `S_(2)` touch each other externally, `C_(2)C_(3)=sqrt(25+(25)/(4))=(5sqrt(5))/(2)=r_(3)-r_(2)` So, `S_(2)` and `S_(3)` touch each other internally. The point of contact `P_(1)` divides `C_(1)C_(2)` internally in the ratio `r_(1) : r_(2) = 2:1` `implies P_(1) -= (-2 ,-1)` The point of contact `P_(2)` divides `C_(2)C_(3)` externally in the ratio `r_(2) : r_(3) = 1:6` `implies P_(2) -= (-4, -1)`. The point of contact `P_(3)` divides `C_(3)C_(1)` externally in the ratio `r_(3) : r_(1) = 3:1` `implies P_(3) -= ( -1,-4)` Area of `Delta P_(1)P_(2)P_(3)= (1)/(2) | {:(-2,-1,1),(-4,-1,1),(-1,-4,1):}| = 3` And area `Delta C_(1)C_(2)C_(3)= (1)/(2) |{:(0,-2,1),(-3,(-1)/(2),1),(2,2,1):}|=(15)/(2)` `:. ("area "(Delta P_(1)P_(2)P_(3)))/("area"(DeltaC_(1)C_(2)C_(3)))=(3)/(15)=2:5` Clearly, `P_(2)(-4,-1)` and `P_(3)(-1,-4)` are images of each other with respect to the line `y=x` |
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