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

If two circles intersect at two distinct points->|r1 – r2| < C1C2 < r1 + r2| r – 2| < 9 + 1 6 < 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

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

3 < r < 7

5 < r < 9

<math> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> </math> < r < 4

0 < r < 3

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Conic Sections Ncert Solutions Maths class 11th Maths Class 11th

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 -

$\frac{1}{2}$ < r < 4

Option 4 -

0 < r < 3

3 Views | Posted 4 weeks ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

4 weeks ago

Correct Option - 1

Detailed Solution:

If two circles intersect at two distinct points

->|r₁ – r₂| < C₁C₂ < r₁ + r₂

| r – 2| < $\sqrt[]{9 + 16}$ < 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

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Slope of axis = $\frac{1}{2}$

$y - 3 = \frac{1}{2} (x - 2)$

⇒ 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)

⇒ $\frac{{(\frac{24}{10})}^{2}}{a^{2}} + \frac{{(\frac{32}{10})}^{2}}{b^{2}} = 1$

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$\frac{144}{25} + \frac{256}{25} \times \frac{1}{2} = a^{2}$

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-> $b^{2} = \frac{2 \times 272}{25}$

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(Latus rectum)²

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