If the two circles x^2+y^2-2x-6y+10-r^2=0 and | vedclass

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(A) The first circle is $x^2+y^2-2x-6y+(10-r^2)=0$. Its center $C_1 = (1, 3)$ and radius $r_1 = \sqrt{1^2+3^2-(10-r^2)} = \sqrt{1+9-10+r^2} = |r|$.The

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$2 < |r| < 8$

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(A) The first circle is $x^2+y^2-2x-6y+(10-r^2)=0$. Its center $C_1 = (1, 3)$ and radius $r_1 = \sqrt{1^2+3^2-(10-r^2)} = \sqrt{1+9-10+r^2} = |r|$.The second circle is $x^2+y^2-8x+2y+8=0$. Its center $C_2 = (4, -1)$ and radius $r_2 = \sqrt{4^2+(-1)^2-8} = \sqrt{16+1-8} = 3$.The distance between the centers $C_1$ and $C_2$ is $d = \sqrt{(4-1)^2+(-1-3)^2} = \sqrt{3^2+(-4)^2} = \sqrt{9+16} = 5$.For two circles to have a common chord of non-zero length,they must intersect at two distinct points. This occurs if $|r_1 - r_2| Substituting the values: $| |r| - 3 | Case $1$: $|r| + 3 > 5 \implies |r| > 2$.Case $2$: $| |r| - 3 | Combining these,we get $2 < |r| < 8$.

If the two circles $x^2+y^2-2x-6y+10-r^2=0$ and $x^2+y^2-8x+2y+8=0$ have a common chord of non-zero length,then

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