Prove that length of a diameter of circle $$ x^{2}+y^{2}=q^{2} $$ is $$ 2 a $$

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To prove that the length of a diameter of the circle $$ x^{2} + y^{2} = q^{2} $$ is $$ 2q $$, let's follow these steps:

### 1. **Identify the Standard Form of a Circle**

The general equation of a circle in Cartesian coordinates is:

$$
(x - h)^2 + (y - k)^2 = r^{2}
$$

where:
- $$(h, k)$$ is the **center** of the circle.
- $$r$$ is the **radius** of the circle.

### 2. **Rewrite the Given Equation in Standard Form**

The given equation is:

$$
x^{2} + y^{2} = q^{2}
$$

Comparing this with the standard form:

$$
(x - 0)^2 + (y - 0)^2 = q^{2}
$$

We can see that:
- The **center** of the circle is at $$(0, 0)$$.
- The **radius** $$ r $$ is $$ q $$.

### 3. **Understand the Relationship Between Radius and Diameter**

The **diameter** of a circle is a straight line passing through the center of the circle, connecting two points on its circumference. The relationship between the radius ($$ r $$) and the diameter ($$ d $$) is:

$$
d = 2r
$$

### 4. **Calculate the Diameter**

Given that the radius $$ r = q $$, the diameter $$ d $$ is:

$$
d = 2q
$$

### 5. **Conclusion**

Therefore, the length of the diameter of the circle $$ x^{2} + y^{2} = q^{2} $$ is $$ 2q $$.

$\boxed{2q}$
The diameter of the circle $$ x^{2} + y^{2} = q^{2} $$ is $$ 2q $$.

Prove that length of a diameter of circle \( x^{2}+y^{2}=q^{2} \) is \( 2 a \)

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