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  4. If the circles x2 + y2 - 16x-20y + 164 = r2 and (x-4)2 + (y - 7)2 = 36 intersect at two distinct points, then:

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Q. 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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Solution:

$x^2 + y^2 - 16x - 20y + 164 = r^2$
$A(8,10), R_1 = r$
$(x - 4)^2 + (y - 7)^2 = 36 $
$B(4,7), R_2 = 6$
$|R_1 - R_2| < AB < R_1 + R_2$
$\Rightarrow \; 1 < r < 11$