If a circle C, whose radius is 3, touches the | vedclass

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(D) The given circle is $x^2 + y^2 + 2x - 4y - 4 = 0.$ Its center is $O_1 = (-1, 2)$ and its radius $r_1 = \sqrt{(-1)^2 + 2^2 - (-4)} = \sqrt{1 + 4 +

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$2\sqrt{5}$

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(D) The given circle is $x^2 + y^2 + 2x - 4y - 4 = 0.$ Its center is $O_1 = (-1, 2)$ and its radius $r_1 = \sqrt{(-1)^2 + 2^2 - (-4)} = \sqrt{1 + 4 + 4} = 3.$ Let the center of circle $C$ be $O_2 = (h, k)$ and its radius be $r_2 = 3.$ Since the circles touch externally at $P(2, 2),$ the point $P$ divides the line segment $O_1O_2$ internally in the ratio $r_1 : r_2 = 3 : 3 = 1 : 1.$ Thus,$P$ is the midpoint of $O_1O_2.$ $(2, 2) = \left( \frac{-1 + h}{2}, \frac{2 + k}{2} \right).$ Solving for $h$ and $k$: $-1 + h = 4 \Rightarrow h = 5$ $2 + k = 4 \Rightarrow k = 2$ So,the center of circle $C$ is $(5, 2).$ The equation of circle $C$ is $(x - 5)^2 + (y - 2)^2 = 3^2,$ which simplifies to $x^2 - 10x + 25 + y^2 - 4y + 4 = 9,$ or $x^2 + y^2 - 10x - 4y + 20 = 0.$ The length of the intercept cut by this circle on the $x-$axis is given by $2\sqrt{g^2 - c},$ where $g = -5$ and $c = 20.$ Length $= 2\sqrt{(-5)^2 - 20} = 2\sqrt{25 - 20} = 2\sqrt{5}.$

If a circle $C,$ whose radius is $3,$ touches the circle $x^2 + y^2 + 2x - 4y - 4 = 0$ externally at the point $(2, 2),$ then the length of the intercept cut by circle $C$ on the $x-$axis is equal to

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