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We are tasked with finding the standard form of the equation of a circle centered at the origin and containing the point \((5, 10)\).

### Standard Equation of a Circle:
The equation of a circle with center at the origin \((0,0)\) is given by:
\[
x^2 + y^2 = r^2
\]
where \(r\) is the radius of the circle.

### Step 1: Find the Radius \(r\)
The radius \(r\) is the distance from the center (origin) to the point \((5, 10)\). We can use the distance formula:
\[
r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
Substituting \((x_1, y_1) = (0, 0)\) and \((x_2, y_2) = (5, 10)\):
\[
r = \sqrt{(5 - 0)^2 + (10 - 0)^2} = \sqrt{5^2 + 10^2} = \sqrt{25 + 100} = \sqrt{125}
\]
Thus, \(r = \sqrt{125}\).

### Step 2: Write the Equation of the Circle
Now that we know \(r = \sqrt{125}\), we substitute \(r^2 = 125\) into the standard form:
\[
x^2 + y^2 = 125
\]

Therefore, the correct equation of the circle is:
\[
x^2 + y^2 = 125
\]

The correct answer is the second option: \(x^2 + y^2 = 125\).

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Would you like further details on this, or do you have any questions?

Here are 5 related questions:
1. How do you find the equation of a circle when it's not centered at the origin?
2. Can you generalize the equation for a circle given any center \((h, k)\)?
3. What is the importance of the distance formula in geometry?
4. How do you determine if a point lies inside, outside, or on the circle?
5. How does the Pythagorean theorem relate to the equation of a circle?

**Tip:** The square of the radius (\(r^2\)) in a circle's equation helps avoid dealing with square roots in the equation itself.

Find the standard form of the equation of the circle. Center at the origin and containing the point (5, 10).

Learn how to find the standard form of a circle's equation, centered at the origin and passing through the point (5, 10). Understand the relationship between the distance formula and the radius of the circle. Suitable for high school students in grades 9-11.

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