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If the circles {x}^{{2}}+{y}^{{2}}-{16}{x}-{20}{y}+{164}={r}^{

Question

If the circles $x^{2} + y^{2} - 16 x - 20 y + 164 = r^{2}$ and $(x - 4)^{2} + (y - 7)^{2} = 36$ intersect at two points then (a) $1 < r < 11$ (b) $r = 11$ (c) $r > 11$ (d) $0 < r < 1$

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Luna Discussed

If the circles

x^{2} + y^{2} - 16 x - 20 y + 164 = r^{2}

and

(x - 4)^{2} + (y - 7)^{2} = 36

intersect at two points then (a)

1 < r < 11

(b)

r = 11

(c)

r > 11

(d)

0 < r < 1

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Question Text	If the circles $x^{2} + y^{2} - 16 x - 20 y + 164 = r^{2}$ and $(x - 4)^{2} + (y - 7)^{2} = 36$ intersect at two points then (a) $1 < r < 11$ (b) $r = 11$ (c) $r > 11$ (d) $0 < r < 1$
Answer Type	Video solution: 1
Upvotes	150