Given,
S1≡(x+1)2+(y+2)2=r2 and S2≡x2+y2−4x−4y+4=0 or S2≡(x−2)2+(y−2)2=22
Now, if two circles intersect at two distinct points then ∣r1−r2∣<C1C2<r1+r2.
⇒∣r−2∣<32+42<r+2
⇒∣r−2∣<5<r+2
⇒∣r−2∣<5andr+2>5
⇒−5<r−2<5andr>3
⇒−3<r<7andr>3
⇒3<r<7
If the circles (x+1)2+(y+2)2=r2 and x2+y2−4x−4y+4=0 intersect at exactly two distinct points, then
Held on 30 Jan 2024 · Verified 6 Jul 2026.
5<r<9
0<r<7
3<r<7
21<r<7
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