If the circles x^2 + y^2 - 16x - 20y + 164 = | vedclass

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(B) The first circle is $x^2 + y^2 - 16x - 20y + 164 = r^2$. Rewriting in standard form: $(x - 8)^2 + (y - 10)^2 = r^2$. The center is $A(8, 10)$ and

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$1 < r < 11$

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(B) The first circle is $x^2 + y^2 - 16x - 20y + 164 = r^2$. Rewriting in standard form: $(x - 8)^2 + (y - 10)^2 = r^2$. The center is $A(8, 10)$ and the radius is $R_1 = r$.The second circle is $(x - 4)^2 + (y - 7)^2 = 36$. The center is $B(4, 7)$ and the radius is $R_2 = 6$.The distance between the centers $A$ and $B$ is $AB = \sqrt{(8 - 4)^2 + (10 - 7)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = 5$.For two circles to intersect at two distinct points,the condition is $|R_1 - R_2| Substituting the values: $|r - 6| From $r + 6 > 5$,we get $r > -1$. Since $r$ is a radius,$r > 0$.From $|r - 6| Combining these,the condition is $1 < r < 11$.

If the circles $x^2 + y^2 - 16x - 20y + 164 = r^2$ and $(x - 4)^2 + (y - 7)^2 = 36$ intersect at two distinct points,then

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