The range of r for which circles (x + 1)2 + (y + 2)2 = r2 and x2 + y2 – 4x – 4y + 4 = 0 coincide at two distinct points
The range of r for which circles (x + 1)2 + (y + 2)2 = r2 and x2 + y2 – 4x – 4y + 4 = 0 coincide at two distinct points
Option 1 -
3 < r < 7
Option 2 -
5 < r < 9
Option 3 -
< r < 4
Option 4 -
0 < r < 3
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1 Answer
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Correct Option - 1
Detailed Solution:If two circles intersect at two distinct points
->|r1 – r2| < C1C2 < r1 + r2
| r – 2| < < r + 2
|r – 2| < 5 and r + 2 > 5
–5 < r 2 < 5 r > 3 … (2)
–3 < r < 7 (1)
From (1) and (2)
3 < r < 7
Similar Questions for you
ae = 2b
Or 4 (1 – e2) = e2
4 = 5e2 ->
x2 – y2 cosec2q = 5
x2 cosec2q + y2 = 5
and
->
1 + sin2q = 7 – 7 sin2q
->8sin2q = 6
->
->

Slope of axis =
⇒ 2y – 6 = x – 2
⇒ 2y – x – 4 = 0
2x + y – 6 = 0
4x + 2y – 12 = 0
α + 1.6 = 4 ⇒ α = 2.4
β + 2.8 = 6 ⇒ β = 3.2
Ellipse passes through (2.4, 3.2)
⇒
&
K
K =
48 = 16 (e2 – 1)
= 2 cos2 θ + 2 sin2 θ + 6 sin θ + 45
= 6 sin θ + 47
for maximum of PA2 + PB2, sin q = 1
then P (1, 2)
Hence P, A & B will lie on a straight line.
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