A rod AB of length 3 units moves vertically with its bottom B always an the circle `x^(2)+y^(2)=25` then the equation of the locus of A is

Options

A. `x^(2)+(y+3)^(2)=25`
B. `(x-3)^(2)+y^(2)=25`
C. `(x+3)^(2)+y^(2)=25`
D. `x^(2)+(y-3)^(2)=25`

Answer

To find the equation of the locus of point A, we start by analyzing the given information:

1. **Understanding the Problem**: We have a rod AB of length 3 units that moves vertically with its bottom point B always on the circle defined by the equation \(x^2 + y^2 = 25\).

2. **Circle Properties**: The equation \(x^2 + y^2 = 25\) represents a circle with a center at the origin (0, 0) and a radius of 5 units.

3. **Coordinates of Point B**: Since point B is always on the circle, we can denote its coordinates as \(B(x_1, y_1)\), where \(x_1^2 + y_1^2 = 25\).

4. **Position of Point A**: The rod AB has a fixed length of 3 units. If point B is at coordinates \((x_1, y_1)\), then point A, which is directly above point B (since the rod moves vertically), will have coordinates \((x_1, y_1 + 3)\).

5. **Finding the Locus of Point A**: To find the locus of point A, we need to express the coordinates of A in terms of the coordinates of B. Since \(y_1 = y - 3\), we can substitute this into the circle's equation:

\[ x_1^2 + (y - 3)^2 = 25 \]

6. **Expanding the Equation**: Now, we expand the equation:

\[ x_1^2 + (y^2 - 6y + 9) = 25 \]

Rearranging gives:

\[ x_1^2 + y^2 - 6y + 9 = 25 \]

\[ x_1^2 + y^2 - 6y - 16 = 0 \]

7. **Expressing in Standard Form**: We can rearrange this to express it in a more recognizable form:

\[ x^2 + (y - 3)^2 = 16 \]

This represents a circle with center at \((0, 3)\) and radius 4.

8. **Final Equation of the Locus**: Therefore, the equation of the locus of point A is:

\[ x^2 + (y - 3)^2 = 16 \]