Question
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 (a) $2<\mathrm{r}<8$(b) $r<2$(c) $r=2$(d) $r>2$
Step 1
The first circle has center at $(1,3)$ and radius $r$. The second circle can be rewritten as $(x-4)^2+(y+1)^2=9$, so it has center at $(4,-1)$ and radius $3$. Show more…
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