A circle C_1 of unit radius lies in the first | vedclass

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(C) The circle $C_1$ has radius $r_1 = 1$ and touches both axes in the first quadrant,so its center is $(1, 1)$. Its equation is $(x-1)^2 + (y-1)^2 =

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$3$

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(C) The circle $C_1$ has radius $r_1 = 1$ and touches both axes in the first quadrant,so its center is $(1, 1)$. Its equation is $(x-1)^2 + (y-1)^2 = 1$.Let the second circle $C_2$ have radius $r$ and center $(r, r)$. Its equation is $(x-r)^2 + (y-r)^2 = r^2$.The common chord of two intersecting circles is a line perpendicular to the line joining their centers. The length of the common chord is maximized when the common chord passes through the center of the smaller circle or when the circles are positioned such that the chord is the diameter of the smaller circle.For two circles $(x-h_1)^2 + (y-k_1)^2 = r_1^2$ and $(x-h_2)^2 + (y-k_2)^2 = r_2^2$,the common chord is the radical axis: $2x(h_2-h_1) + 2y(k_2-k_1) + h_1^2 + k_1^2 - r_1^2 - h_2^2 - k_2^2 + r_2^2 = 0$.Here,$h_1=1, k_1=1, r_1=1$ and $h_2=r, k_2=r, r_2=r$.The radical axis equation is $2x(r-1) + 2y(r-1) + 1 + 1 - 1 - r^2 - r^2 + r^2 = 0$,which simplifies to $2(r-1)(x+y) + 1 - r^2 = 0$.The length of the common chord is maximized when the radical axis passes through the center of the circle with the smaller radius. Setting $x=1, y=1$ in the radical axis equation: $2(r-1)(2) + 1 - r^2 = 0 \implies 4r - 4 + 1 - r^2 = 0 \implies r^2 - 4r + 3 = 0$.Solving for $r$: $(r-3)(r-1) = 0$. Since $r 
eq 1$ (as the circles would be identical),we get $r = 3$.

$A$ circle $C_1$ of unit radius lies in the first quadrant and touches both coordinate axes. The radius of another circle that touches both coordinate axes and intersects $C_1$ such that the common chord is of maximum length is:

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