If the circle x^2 + y^2 + 2x - 4y - k = 0 lie | vedclass

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(C) The given equations of the circles are of the form $x^2 + y^2 + 2x - 4y + C = 0$.These circles are concentric with the center $(-g, -f) = (-1, 2)$

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(C) The given equations of the circles are of the form $x^2 + y^2 + 2x - 4y + C = 0$.These circles are concentric with the center $(-g, -f) = (-1, 2)$.The radius $r$ of a circle $x^2 + y^2 + 2x - 4y + C = 0$ is given by $r = \sqrt{g^2 + f^2 - C} = \sqrt{1 + 4 - C} = \sqrt{5 - C}$.Let the radii of the three circles be $r_1, r_2,$ and $r_3$ respectively.For the first circle,$C_1 = -4$,so $r_1^2 = 5 - (-4) = 9$,which means $r_1 = 3$.For the third circle,$C_3 = -20$,so $r_3^2 = 5 - (-20) = 25$,which means $r_3 = 5$.Since the middle circle lies exactly between the other two,its radius $r_2$ must be the arithmetic mean of $r_1$ and $r_3$.$r_2 = \frac{r_1 + r_3}{2} = \frac{3 + 5}{2} = 4$.Now,for the middle circle,$r_2^2 = 5 - (-k) = 5 + k$.Since $r_2 = 4$,$r_2^2 = 16$.Therefore,$5 + k = 16$,which gives $k = 11$.

If the circle $x^2 + y^2 + 2x - 4y - k = 0$ lies exactly between the circles $x^2 + y^2 + 2x - 4y - 4 = 0$ and $x^2 + y^2 + 2x - 4y - 20 = 0$,then $k = \dots$

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