If `x=3+2cos theta, y=5+2sin theta` then the locus of the point (x,y) is a circle with centre and radius

Options

A. (2,5),-2
B. (3,5),2
C. (5,7),6
D. (8,11),4

Answer

To find the locus of the point (x, y) given by the equations \( x = 3 + 2 \cos \theta \) and \( y = 5 + 2 \sin \theta \), we will follow these steps:

### Step 1: Express \(\cos \theta\) and \(\sin \theta\) in terms of \(x\) and \(y\)

From the equations provided:
- For \(x\):
\[ x - 3 = 2 \cos \theta \implies \cos \theta = \frac{x - 3}{2} \]

- For \(y\):
\[ y - 5 = 2 \sin \theta \implies \sin \theta = \frac{y - 5}{2} \]

### Step 2: Use the Pythagorean identity

We know that:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]

Substituting the expressions we found for \(\sin \theta\) and \(\cos \theta\):
\[ \left(\frac{y - 5}{2}\right)^2 + \left(\frac{x - 3}{2}\right)^2 = 1 \]

### Step 3: Simplify the equation

Squaring both terms:
\[ \frac{(y - 5)^2}{4} + \frac{(x - 3)^2}{4} = 1 \]

Multiplying the entire equation by 4 to eliminate the denominators:
\[ (y - 5)^2 + (x - 3)^2 = 4 \]

### Step 4: Identify the center and radius of the circle

The equation \((y - 5)^2 + (x - 3)^2 = 4\) is in the standard form of a circle:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
where \((h, k)\) is the center and \(r\) is the radius.

From our equation:
- The center \((h, k)\) is \((3, 5)\)
- The radius \(r\) is \(\sqrt{4} = 2\)

### Final Answer

Thus, the locus of the point \((x, y)\) is a circle with:
- Center: \((3, 5)\)
- Radius: \(2\)

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