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

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

1. A

$1

2. B

$r>11$

3. C

$0

4. D

$r=11$

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### Solution:

Given equation of circle is ${x}^{2}+{y}^{2}-16x-20y+164={r}^{2}⇒\left(x-8{\right)}^{2}+\left(y-10{\right)}^{2}={r}^{2}$

${C}_{1}=\left(8,10\right)$ and ${R}_{1}=r$

for the second circle ${C}_{2}$=(4,7),${R}_{2}$=6

now, ${C}_{1}{C}_{2}=\sqrt{\left(8-4{\right)}^{2}+\left(10-7{\right)}^{2}}=\sqrt{25}=5$

Since the circles intersect at two distinct points So,

${R}_{1}+{R}_{2}>{C}_{1}{C}_{2}>\left|{R}_{1}-{R}_{2}\right|$

$⇒r+6>5>|r-6|⇒1

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