O(0,0) and A(1,0) are the centers of two unit | vedclass

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(A) The centers of $C_1$ and $C_2$ are $O(0,0)$ and $A(1,0)$ with radius $r=1$.$C_3$ is a unit circle passing through $O(0,0)$ and $A(1,0)$. Let its c

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$\sqrt{3}x - y + 2 = 0$

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(A) The centers of $C_1$ and $C_2$ are $O(0,0)$ and $A(1,0)$ with radius $r=1$.$C_3$ is a unit circle passing through $O(0,0)$ and $A(1,0)$. Let its center be $(h, k)$.Since it passes through $(0,0)$,$h^2 + k^2 = 1^2 = 1$.Since it passes through $(1,0)$,$(h-1)^2 + k^2 = 1^2 = 1$.Subtracting the equations: $h^2 - (h-1)^2 = 0 \implies h^2 - (h^2 - 2h + 1) = 0 \implies 2h - 1 = 0 \implies h = 1/2$.Then $(1/2)^2 + k^2 = 1 \implies k^2 = 3/4 \implies k = \sqrt{3}/2$ (since center is above $X$-axis).So $C_3$ has center $(1/2, \sqrt{3}/2)$ and radius $1$.$C_1$ has center $(0,0)$ and radius $1$. The line $L: ax + by + c = 0$ is tangent to $C_1$ if $|c|/\sqrt{a^2+b^2} = 1 \implies c^2 = a^2 + b^2$.It is tangent to $C_3$ if $|a/2 + b\sqrt{3}/2 + c|/\sqrt{a^2+b^2} = 1 \implies |a + b\sqrt{3} + 2c| = 2\sqrt{a^2+b^2}$.Substituting $c^2 = a^2+b^2$,we test the options. For option $A$: $\sqrt{3}x - y + 2 = 0$,$a=\sqrt{3}, b=-1, c=2$. $a^2+b^2 = 3+1=4=c^2$. Tangent to $C_1$.For $C_3$: $|\sqrt{3}(1/2) - (\sqrt{3}/2) + 2| = |2| = 2$. $\sqrt{a^2+b^2} = 2$. $2/2 = 1$. Tangent to $C_3$.Checking intersection with $C_2$ (center $(1,0)$,radius $1$): Distance from $(1,0)$ to $\sqrt{3}x - y + 2 = 0$ is $|\sqrt{3}(1) - 0 + 2|/\sqrt{3+1} = |\sqrt{3}+2|/2 \approx 3.73/2 > 1$. It does not intersect $C_2$.

$O(0,0)$ and $A(1,0)$ are the centers of two unit circles $C_1$ and $C_2$ respectively. $C_3$ is also a unit circle having its center above the $X$-axis and passing through $O$ and $A$. The equation of the common tangent to $C_1$ and $C_3$ which does not intersect the circle $C_2$ is

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