A circle cuts an intercept of length 12 units from the x-axis and its centre lies at the origin. Its equation is ............

Answer

To find the equation of a circle that cuts an intercept of length 12 units from the x-axis and has its center at the origin, we can follow these steps:

### Step 1: Understand the problem
The circle has its center at the origin (0, 0) and intercepts the x-axis at two points. The total length of this intercept is given as 12 units.

### Step 2: Determine the points of intersection
Since the intercept is 12 units long, the two points where the circle intersects the x-axis can be found. Let's denote these points as A and B. The distance from A to B is 12 units.

If we let the coordinates of point A be (-6, 0) and point B be (6, 0), we can see that the distance between these two points is:
\[ AB = |6 - (-6)| = 12 \text{ units} \]

### Step 3: Find the radius of the circle
The radius of the circle can be determined by finding the distance from the center (the origin) to either point A or point B. The distance from the origin (0, 0) to point A (-6, 0) is:
\[ OA = \sqrt{(-6 - 0)^2 + (0 - 0)^2} = \sqrt{36} = 6 \]
Thus, the radius \( r \) of the circle is 6 units.

### Step 4: Write the equation of the circle
The standard form of the equation of a circle with center at the origin (0, 0) and radius \( r \) is given by:
\[ x^2 + y^2 = r^2 \]
Substituting \( r = 6 \):
\[ x^2 + y^2 = 6^2 \]
\[ x^2 + y^2 = 36 \]

### Step 5: Rearranging the equation
We can rearrange the equation to the form:
\[ x^2 + y^2 - 36 = 0 \]

### Final Answer
The equation of the circle is:
\[ x^2 + y^2 - 36 = 0 \]

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