If one common tangent of the two circles x^2 | vedclass

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(A) The equation of the tangent to the circle $x^2 + y^2 = 4$ at the point $(\sqrt{3}, 1)$ is given by $T = 0$,which is $x x_1 + y y_1 = r^2$.Substitu

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$\frac{1}{4}$

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(A) The equation of the tangent to the circle $x^2 + y^2 = 4$ at the point $(\sqrt{3}, 1)$ is given by $T = 0$,which is $x x_1 + y y_1 = r^2$.Substituting the point $(\sqrt{3}, 1)$,we get $\sqrt{3}x + 1y = 4$,or $\sqrt{3}x + y - 4 = 0$.This line is also a tangent to the second circle $x^2 + (y - 3)^2 = \lambda$. The center of this circle is $(0, 3)$ and its radius is $\sqrt{\lambda}$.The perpendicular distance from the center $(0, 3)$ to the line $\sqrt{3}x + y - 4 = 0$ must be equal to the radius $\sqrt{\lambda}$.Distance $d = \frac{|\sqrt{3}(0) + 1(3) - 4|}{\sqrt{(\sqrt{3})^2 + 1^2}} = \frac{|3 - 4|}{\sqrt{3 + 1}} = \frac{|-1|}{2} = \frac{1}{2}$.Since $d^2 = \lambda$,we have $\lambda = (\frac{1}{2})^2 = \frac{1}{4}$.

$A. (7, 2)$	$1. -4\sqrt{2}$
$B. (0, 1/\sqrt{2})$	$2. -8$
$C. (1, -1)$	$3. 4$
$D. (3, \sqrt{2})$	$4. 0$
	$5. -2\sqrt{2}$

If one common tangent of the two circles $x^2 + y^2 = 4$ and $x^2 + (y - 3)^2 = \lambda, \lambda > 0$ passes through the point $(\sqrt{3}, 1)$,then the possible value of $\lambda$ is

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$B. (0, 1/\sqrt{2})$ $2. -8$
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