Equation of circle throught origin, having radius 5 and abscissa of centre is (-3) .

Options

A. `x^(2) + y^(2) pm 8x + 6y = 0 `
B. `x^(2) + y^(2) + 6x pm 8y -25 = 0 `
C. `x^(2) + y^(2) + 6x pm 8y = 0 `
D. `x^(2) + y^(2) + 3x pm 4y = 0 `

Answer

To find the equation of a circle that passes through the origin, has a radius of 5, and has an abscissa (x-coordinate) of the center as -3, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Center of the Circle**:
- The center of the circle (h, k) is given that the abscissa (h) is -3.
- Since the y-coordinate (k) is not provided, we will denote it as k.
- Therefore, the center of the circle is (-3, k).

2. **Use the Radius**:
- The radius of the circle is given as 5.
- The circle passes through the origin (0, 0).

3. **Apply the Distance Formula**:
- The distance from the center (-3, k) to the origin (0, 0) must equal the radius (5).
- Using the distance formula:
\[ \sqrt{(0 - (-3))^2 + (0 - k)^2} = 5 \]
- This simplifies to:
\[ \sqrt{(3)^2 + (-k)^2} = 5 \]
- Which further simplifies to:
\[ \sqrt{9 + k^2} = 5 \]

4. **Square Both Sides**:
- To eliminate the square root, square both sides:
\[ 9 + k^2 = 25 \]

5. **Solve for k**:
- Rearranging gives:
\[ k^2 = 25 - 9 \]
\[ k^2 = 16 \]
- Taking the square root of both sides gives:
\[ k = \pm 4 \]

6. **Determine the Possible Centers**:
- The possible centers of the circle are (-3, 4) and (-3, -4).

7. **Write the Equation of the Circle**:
- The standard form of the equation of a circle is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
- Substituting h = -3, k = 4 (for one possible center):
\[ (x + 3)^2 + (y - 4)^2 = 5^2 \]
\[ (x + 3)^2 + (y - 4)^2 = 25 \]
- For the other center (-3, -4):
\[ (x + 3)^2 + (y + 4)^2 = 25 \]

### Final Answer:
The equations of the circles are:
1. \((x + 3)^2 + (y - 4)^2 = 25\)
2. \((x + 3)^2 + (y + 4)^2 = 25\)