If the centre of the circle passing through the origin is `(3,4)`.
then the intercept cut off by the circle on x-axis and y-axis respectively are

Options

A. 3 unit and 4 unit
B. 6 unit and 4 unit
C. 3 unit and 8 unit
D. 6 unit and 8 unit

Explanation

To find the intercepts cut off by the circle on the x-axis and y-axis, we will follow these steps:

### Step 1: Write the equation of the circle
The general equation of a circle with center \((h, k)\) and radius \(r\) is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Given the center of the circle is \((3, 4)\), we can write the equation as:
\[ (x - 3)^2 + (y - 4)^2 = r^2 \]

### Step 2: Determine the radius \(r\)
Since the circle passes through the origin \((0, 0)\), we can substitute these coordinates into the equation to find \(r\):
\[ (0 - 3)^2 + (0 - 4)^2 = r^2 \]
Calculating this gives:
\[ 9 + 16 = r^2 \implies r^2 = 25 \implies r = 5 \]

### Step 3: Substitute \(r\) back into the circle's equation
Now that we have \(r^2 = 25\), we can write the complete equation of the circle:
\[ (x - 3)^2 + (y - 4)^2 = 25 \]

### Step 4: Find the x-intercept
To find the x-intercept, set \(y = 0\) in the equation of the circle:
\[ (x - 3)^2 + (0 - 4)^2 = 25 \]
This simplifies to:
\[ (x - 3)^2 + 16 = 25 \]
Subtracting 16 from both sides gives:
\[ (x - 3)^2 = 9 \]
Taking the square root:
\[ x - 3 = 3 \quad \text{or} \quad x - 3 = -3 \]
Thus:
\[ x = 6 \quad \text{or} \quad x = 0 \]
The x-intercepts are \(6\) and \(0\).

### Step 5: Find the y-intercept
To find the y-intercept, set \(x = 0\) in the equation of the circle:
\[ (0 - 3)^2 + (y - 4)^2 = 25 \]
This simplifies to:
\[ 9 + (y - 4)^2 = 25 \]
Subtracting 9 from both sides gives:
\[ (y - 4)^2 = 16 \]
Taking the square root:
\[ y - 4 = 4 \quad \text{or} \quad y - 4 = -4 \]
Thus:
\[ y = 8 \quad \text{or} \quad y = 0 \]
The y-intercepts are \(8\) and \(0\).

### Final Result
The intercepts cut off by the circle on the x-axis and y-axis are:
- X-intercept: \(6\)
- Y-intercept: \(8\)

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