Circle Ques 31
- If the two circles $(x-1)^{2}+(y-3)^{2}=r^{2}$ and $x^{2}+y^{2}-8 x+2 y+8=0$ intersect in two distinct points, then
$(1989,2 M)$
(a) $2<r<8$
(b) $r<2$
(c) $r=2$
(d) $r>2$
Show Answer
Answer:
Correct Answer: 31.(a)
Solution:
Formula:
Common Tangents of Two Circles
- As, the two circles intersect in two distinct points.
$\Rightarrow$ Distance between centres lies between $\left|r _1-r _2\right|$ and $\left|r _1+r _2\right|$
$ \begin{array}{lc} \text { i.e. } & |r-3|<\sqrt{(4-1)^{2}+(-1-3)^{2}}<|r+3| \\ \Rightarrow & |r-3|<5<|r+3| \Rightarrow r<8 \text { or } r>2 \\ \therefore & 2<r<8 \end{array} $
BETA
