Circle 2 Question 4
5. 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 distinct points, then
(2019 Main, 9 Jan II)
(a) $0<r<1$
(b) $r>11$
(c) $1<r<11$
(d) $r=11$
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Answer:
Correct Answer: 5. (b)
Solution:
- Circle I is $x^{2}+y^{2}-16 x-20 y+164=r^{2}$
$\Rightarrow \quad(x-8)^{2}+(y-10)^{2}=r^{2}$
$\Rightarrow C _1(8,10)$ is the centre of Istcircle and $r _1=r$ is its radius Circle II is $(x-4)^{2}+(y-7)^{2}=36$ $\Rightarrow C _2(4,7)$ is the centre of 2 nd circle and $r _2=6$ is its radius.
Two circles intersect if $\left|r _1-r _2\right|<C _1 C _2<r _1+r _2$
$\Rightarrow|r-6|<\sqrt{(8-4)^{2}+(10-7)^{2}}<r+6$
$\Rightarrow|r-6|<\sqrt{16+9}<r+6$
$\Rightarrow|r-6|<5<r+6$
Now as, $5<r+6$ always, we have to solve only
$$ |r-6|<5 \Rightarrow-5<r-6<5 $$
$\Rightarrow \quad 6-5<r<5+6 \Rightarrow 1<r<11$
