If the parametric of form of a circle is given by
(a) `x=-4+5costhetaandy=-3+5sintheta`
(b) `x=acosalpha+bsinalphaandy=asinalpha-bcosalpha` find its cartesian form.

Answer

To convert the given parametric equations of a circle into Cartesian form, we will follow a systematic approach for both parts of the question.

### Part (a)

Given the parametric equations:
1. \( x = -4 + 5 \cos \theta \)
2. \( y = -3 + 5 \sin \theta \)

**Step 1: Rearranging the equations**

Rearranging the equation for \(x\):
\[ x + 4 = 5 \cos \theta \]
Rearranging the equation for \(y\):
\[ y + 3 = 5 \sin \theta \]

**Step 2: Squaring both equations**

Now, square both sides of the rearranged equations:
\[ (x + 4)^2 = (5 \cos \theta)^2 \quad \Rightarrow \quad (x + 4)^2 = 25 \cos^2 \theta \]
\[ (y + 3)^2 = (5 \sin \theta)^2 \quad \Rightarrow \quad (y + 3)^2 = 25 \sin^2 \theta \]

**Step 3: Expanding the squared equations**

Expanding both equations:
\[ x^2 + 8x + 16 = 25 \cos^2 \theta \quad \text{(1)} \]
\[ y^2 + 6y + 9 = 25 \sin^2 \theta \quad \text{(2)} \]

**Step 4: Adding the equations**

Now, add equations (1) and (2):
\[ (x^2 + 8x + 16) + (y^2 + 6y + 9) = 25 \cos^2 \theta + 25 \sin^2 \theta \]
Using the identity \( \cos^2 \theta + \sin^2 \theta = 1 \):
\[ x^2 + y^2 + 8x + 6y + 25 = 25 \]

**Step 5: Simplifying the equation**

Simplifying the equation:
\[ x^2 + y^2 + 8x + 6y = 0 \]

This is the Cartesian form for part (a).

### Part (b)

Given the parametric equations:
1. \( x = a \cos \alpha + b \sin \alpha \)
2. \( y = a \sin \alpha - b \cos \alpha \)

**Step 1: Squaring both equations**

Square both sides of the equations:
\[ x^2 = (a \cos \alpha + b \sin \alpha)^2 \quad \Rightarrow \quad x^2 = a^2 \cos^2 \alpha + b^2 \sin^2 \alpha + 2ab \sin \alpha \cos \alpha \quad \text{(3)} \]
\[ y^2 = (a \sin \alpha - b \cos \alpha)^2 \quad \Rightarrow \quad y^2 = a^2 \sin^2 \alpha + b^2 \cos^2 \alpha - 2ab \sin \alpha \cos \alpha \quad \text{(4)} \]

**Step 2: Adding the equations**

Now, add equations (3) and (4):
\[ x^2 + y^2 = (a^2 \cos^2 \alpha + b^2 \sin^2 \alpha + 2ab \sin \alpha \cos \alpha) + (a^2 \sin^2 \alpha + b^2 \cos^2 \alpha - 2ab \sin \alpha \cos \alpha) \]

**Step 3: Simplifying the equation**

Combine like terms:
\[ x^2 + y^2 = a^2 (\cos^2 \alpha + \sin^2 \alpha) + b^2 (\sin^2 \alpha + \cos^2 \alpha) \]
Using the identity \( \cos^2 \alpha + \sin^2 \alpha = 1 \):
\[ x^2 + y^2 = a^2 + b^2 \]

This is the Cartesian form for part (b).

### Summary of Results
- For part (a): \( x^2 + y^2 + 8x + 6y = 0 \)
- For part (b): \( x^2 + y^2 = a^2 + b^2 \)