The circle passing through origin and making intercepts 6 and -4 onx and y-axes respectively has the centre

Options

A. (3,-2)
B. (-2,4)
C. (8,-4)
D. Both 1 and 2

Answer

To find the center of the circle that passes through the origin and makes intercepts of 6 on the x-axis and -4 on the y-axis, we can follow these steps:

### Step 1: Understand the Circle's Equation
The general equation of a circle can be represented as:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
where \((h, k)\) is the center of the circle and \(r\) is the radius.

### Step 2: Identify the Intercepts
The x-intercept is given as 6, which means the circle intersects the x-axis at the point (6, 0). The y-intercept is given as -4, which means the circle intersects the y-axis at the point (0, -4).

### Step 3: Find the Center
Since the circle passes through the origin (0, 0) and has intercepts at (6, 0) and (0, -4), we can find the center of the circle. The center \((h, k)\) can be calculated as the midpoint of the intercepts.

- The x-intercept (6, 0) and the y-intercept (0, -4) can be used to find the center:
\[ h = \frac{x_1 + x_2}{2} = \frac{6 + 0}{2} = 3 \]
\[ k = \frac{y_1 + y_2}{2} = \frac{0 + (-4)}{2} = -2 \]

Thus, the center of the circle is \((3, -2)\).

### Step 4: Conclusion
The center of the circle that passes through the origin and makes intercepts of 6 and -4 on the x and y axes respectively is:
\[ \text{Center} = (3, -2) \]

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