Inter Maths-2B - Circles - 7 marks important problems Exercise 1(e) II Q 3

Direct Common Tangents of Two Circles

Interactive Tutorial with Step-by-Step Solution

Problem Statement

Find the direct common tangents of the circles:

Circle 1: x² + y² + 22x - 4y - 100 = 0
Circle 2: x² + y² - 22x + 4y + 100 = 0

Step 1: Convert to Standard Form

Circle 1: x² + y² + 22x - 4y - 100 = 0

Completing the square:

Group terms: (x² + 22x) + (y² - 4y) = 100

For x terms: x² + 22x + 121 = (x + 11)² - 121
For y terms: y² - 4y + 4 = (y - 2)² - 4

(x + 11)² - 121 + (y - 2)² - 4 = 100
(x + 11)² + (y - 2)² = 225

Result: Center C₁(-11, 2), radius r₁ = 15

Circle 2: x² + y² - 22x + 4y + 100 = 0

Group terms: (x² - 22x) + (y² + 4y) = -100

For x terms: x² - 22x + 121 = (x - 11)² - 121
For y terms: y² + 4y + 4 = (y + 2)² - 4

(x - 11)² - 121 + (y + 2)² - 4 = -100
(x - 11)² + (y + 2)² = 25

Result: Center C₂(11, -2), radius r₂ = 5

Step 2: Check Circle Positions

Distance between centers:
d = √[(11-(-11))² + (-2-2)²]
d = √[22² + (-4)²] = √[484 + 16] = √500 = 10√5 ≈ 22.36

Since d = 10√5 > r₁ + r₂ = 20, the circles are external and separate.

Therefore, direct common tangents exist.

Step 3: Set Up Distance Equations

Let the tangent line be: ax + by + c = 0

For direct common tangents:

Distance from C₁(-11, 2) = 15
Distance from C₂(11, -2) = 5

|-11a + 2b + c|/√(a² + b²) = 15
|11a - 2b + c|/√(a² + b²) = 5

Step 4: Solve the System

For direct tangents, both expressions have the same sign:

-11a + 2b + c = ±15√(a² + b²)
11a - 2b + c = ±5√(a² + b²)

Solving the algebra (subtracting and adding equations):

96a² - 44ab - 21b² = 0
96(a/b)² - 44(a/b) - 21 = 0

Using quadratic formula: a/b = 3/4 or a/b = -7/12

Step 5: Find the Tangent Lines

For a/b = 3/4:

Let a = 3, b = 4
Using distance conditions: c = -50
First tangent: 3x + 4y - 50 = 0

For a/b = -7/12:

Let a = 7, b = -24 (scaled appropriately)
Using distance conditions: c = -250
Second tangent: 7x - 24y - 250 = 0

Step 6: Verification

For 3x + 4y - 50 = 0:

Distance from C₁(-11, 2): |3(-11) + 4(2) - 50|/5 = 75/5 = 15 ✓
Distance from C₂(11, -2): |3(11) + 4(-2) - 50|/5 = 25/5 = 5 ✓

For 7x - 24y - 250 = 0:

Distance from C₁(-11, 2): |7(-11) - 24(2) - 250|/25 = 375/25 = 15 ✓
Distance from C₂(11, -2): |7(11) - 24(-2) - 250|/25 = 125/25 = 5 ✓

Final Answer

The direct common tangents are:

1. 3x + 4y - 50 = 0
2. 7x - 24y - 250 = 0

Interactive Visualization

Legend:

🔴 Circle 1: Center(-11, 2), r=15

🔵 Circle 2: Center(11, -2), r=5

🟢 Direct Common Tangents

ipe 2a problem 4

Complex Number Identity Proof

Problem Statement

If \( n \) is an integer, show that:

\[ (1 + \cosθ + i\sinθ)^n + (1 + \cosθ - i\sinθ)^n = 2^{n+1}\cos^n\left(\frac{θ}{2}\right)\cos\left(\frac{nθ}{2}\right) \]

Proof

Step 1: Rewrite the Expression

Let's rewrite \( 1 + \cosθ \) using a trigonometric identity:

\[ 1 + \cosθ = 2\cos^2\left(\frac{θ}{2}\right) \]

Step 2: Express in Polar Form

Now express the complex numbers in terms of half-angle:

\[ 1 + \cosθ + i\sinθ = 2\cos\left(\frac{θ}{2}\right)\left[\cos\left(\frac{θ}{2}\right) + i\sin\left(\frac{θ}{2}\right)\right] \] \[ 1 + \cosθ - i\sinθ = 2\cos\left(\frac{θ}{2}\right)\left[\cos\left(\frac{θ}{2}\right) - i\sin\left(\frac{θ}{2}\right)\right] \]

Step 3: Raise to Power n

Using De Moivre's Theorem:

\[ (1 + \cosθ + i\sinθ)^n = 2^n\cos^n\left(\frac{θ}{2}\right)\left[\cos\left(\frac{nθ}{2}\right) + i\sin\left(\frac{nθ}{2}\right)\right] \] \[ (1 + \cosθ - i\sinθ)^n = 2^n\cos^n\left(\frac{θ}{2}\right)\left[\cos\left(\frac{nθ}{2}\right) - i\sin\left(\frac{nθ}{2}\right)\right] \]

Step 4: Add the Two Expressions

\[ \text{Sum} = 2^n\cos^n\left(\frac{θ}{2}\right)\left[2\cos\left(\frac{nθ}{2}\right)\right] \] \[ = 2^{n+1}\cos^n\left(\frac{θ}{2}\right)\cos\left(\frac{nθ}{2}\right) \]

Final Result

\[ \boxed{(1 + \cosθ + i\sinθ)^n + (1 + \cosθ - i\sinθ)^n = 2^{n+1}\cos^n\left(\frac{θ}{2}\right)\cos\left(\frac{nθ}{2}\right)} \]

Special Cases Verification

Case n = 1:

\[ \text{LHS} = (1 + \cosθ + i\sinθ) + (1 + \cosθ - i\sinθ) = 2(1 + \cosθ) \] \[ \text{RHS} = 2^{2}\cos\left(\frac{θ}{2}\right)\cos\left(\frac{θ}{2}\right) = 4\cos^2\left(\frac{θ}{2}\right) = 2(1 + \cosθ) \]

Case θ = 0:

\[ \text{LHS} = (2)^n + (2)^n = 2^{n+1} \] \[ \text{RHS} = 2^{n+1}\cos^n(0)\cos(0) = 2^{n+1} \]

Case θ = π:

\[ \text{LHS} = (0 + i\sinπ)^n + (0 - i\sinπ)^n = 0 \] \[ \text{RHS} = 2^{n+1}\cos^n\left(\frac{π}{2}\right)\cos\left(\frac{nπ}{2}\right) = 0 \]