If the circles x^2+y^2-2x+4y+c=0 and x^2+y^2+ | vedclass

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(C) The given equations of the circles are:$x^2+y^2-2x+4y+c=0$ ...$(i)$$x^2+y^2+2x-4y+c=0$ ...(ii)For circle $(i)$,the center $C_1 = (1, -2)$ and radi

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$0 < c < 5$

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(C) The given equations of the circles are:$x^2+y^2-2x+4y+c=0$ ...$(i)$$x^2+y^2+2x-4y+c=0$ ...(ii)For circle $(i)$,the center $C_1 = (1, -2)$ and radius $r_1 = \sqrt{1^2 + (-2)^2 - c} = \sqrt{5-c}$.For circle (ii),the center $C_2 = (-1, 2)$ and radius $r_2 = \sqrt{(-1)^2 + 2^2 - c} = \sqrt{5-c}$.Two circles have four common tangents if they are separate,which implies the distance between their centers is greater than the sum of their radii: $d(C_1, C_2) > r_1 + r_2$.The distance between centers $C_1(1, -2)$ and $C_2(-1, 2)$ is $d = \sqrt{(-1-1)^2 + (2 - (-2))^2} = \sqrt{(-2)^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$.So,$2\sqrt{5} > \sqrt{5-c} + \sqrt{5-c} = 2\sqrt{5-c}$.Dividing by $2$,we get $\sqrt{5} > \sqrt{5-c}$.Squaring both sides,$5 > 5-c$,which implies $c > 0$.Also,for the radius to be defined,$5-c > 0$,so $c Combining these,we get $0 < c < 5$.

If the circles $x^2+y^2-2x+4y+c=0$ and $x^2+y^2+2x-4y+c=0$ have four common tangents,then

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