The circle Chas equation $$ x^{2}+y^{2}-10 x+6 y+30=0 $$ Find the radius of $$ C $$

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To find the radius of the circle given by the equation

$$
x^{2}+y^{2}-10 x+6 y+30=0,
$$

we first need to rewrite the equation in standard form. The standard form of a circle's equation is

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

where $$(h, k)$$ is the center of the circle and $$r$$ is the radius.

### Step 1: Rearranging the equation

We start by rearranging the given equation:

$$
x^{2} - 10x + y^{2} + 6y + 30 = 0.
$$

### Step 2: Completing the square

Next, we complete the square for the $$x$$ and $$y$$ terms.

**For $$x$$:**
- Take the coefficient of $$x$$, which is $$-10$$, halve it to get $$-5$$, and square it to get $$25$$.
- Thus, we can rewrite $$x^{2} - 10x$$ as:

$$
(x - 5)^{2} - 25.
$$

**For $$y$$:**
- Take the coefficient of $$y$$, which is $$6$$, halve it to get $$3$$, and square it to get $$9$$.
- Thus, we can rewrite $$y^{2} + 6y$$ as:

$$
(y + 3)^{2} - 9.
$$

### Step 3: Substitute back into the equation

Now substituting these back into the equation gives:

$$
(x - 5)^{2} - 25 + (y + 3)^{2} - 9 + 30 = 0.
$$

Simplifying this:

$$
(x - 5)^{2} + (y + 3)^{2} - 4 = 0.
$$

### Step 4: Rearranging to standard form

Rearranging gives:

$$
(x - 5)^{2} + (y + 3)^{2} = 4.
$$

### Step 5: Identifying the radius

From the standard form, we can see that:

- The center of the circle is $$(5, -3)$$.
- The radius $$r$$ is given by the square root of the right side:

$$
r = \sqrt{4} = 2.
$$

Thus, the radius of the circle $$C$$ is

$$
\boxed{2}.
$$
The radius of circle $$ C $$ is 2.

The circle Chas equation \[ x^{2}+y^{2}-10 x+6 y+30=0 \] Find the radius of \( C \)

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