Question

In Problems 17-22, find the center and radius of the circle with the given equation. x^{2}+y^{2}-10 x+10 y=0

Short answer

The circle's center is (5, -5) and the radius is \(5\sqrt{2}\).

Step-by-step solution

Identify the Circle Equation

The given equation is \(x^2 + y^2 - 10x + 10y = 0\). We need to rewrite it in the form of the standard circle equation \((x-h)^2 + (y-k)^2 = r^2\) to identify the center \((h, k)\) and the radius \(r\).

Complete the Square for x-terms

To complete the square for the \(x\)-terms, we take \(x^2 - 10x\). Add and subtract \(\left(\frac{-10}{2}\right)^2 = 25\) inside the expression. Thus, \(x^2 - 10x = (x - 5)^2 - 25\).

Complete the Square for y-terms

Similarly, for the \(y\)-terms, \(y^2 + 10y\), we add and subtract \(\left(\frac{10}{2}\right)^2 = 25\). This transforms \(y^2 + 10y\) into \((y + 5)^2 - 25\).

Rewrite the Equation

Substitute the completed square forms in the equation: \((x - 5)^2 - 25 + (y + 5)^2 - 25 = 0\). This simplifies to \((x - 5)^2 + (y + 5)^2 = 50\).

Identify the Center and Radius

Comparing with the standard circle equation \((x-h)^2 + (y-k)^2 = r^2\), we find that \(h = 5\), \(k = -5\), and \(r^2 = 50\). Therefore, the center is \((5, -5)\), and the radius \(r = \sqrt{50} = 5\sqrt{2}\).

Key concepts

completing the square

Completing the square is an important technique used in algebra to transform quadratic equations into a more manageable form. This method is crucial for solving equations involving circles, especially when the equation is not initially in standard form.

The process involves transforming a quadratic expression into a perfect square trinomial, which is an expression of the form

• \((x-h)^2\).

To complete the square for an expression like \(x^2 + bx\), we follow these steps:

• Take half of the coefficient of \(x\), which is \(b/2\).
• Square it to get \((b/2)^2\).
• Add and subtract this square inside the equation to maintain balance.

This approach is applied similarly to \(y^2 + dy\) terms. Completing the square allows for the transformation of an equation to easily identify the circle's properties when analyzing its equation.

standard form of a circle

The standard form of a circle's equation is essential for easily discerning the circle's center and radius. This form is written as \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) represents the circle's center and \(r\) the radius.

When an equation is given but not in standard form, as was the case in our original equation \(x^2 + y^2 - 10x + 10y = 0\), you'll need to rearrange it.
By completing the square for both \(x\) and \(y\) terms, you can rewrite the equation to fit this standard format. Once in this form, identifying the circle's characteristics becomes straightforward. Transforming equations into their standard form is a fundamental approach in geometry and particularly useful when graphing or solving for specific points on the circle.

center of a circle

The center of a circle is a crucial element and provides a fixed point from which all the circle's points are equidistant. In the equation of a circle, the center is given by the coordinates \((h, k)\).

In the standard form, \((x-h)^2 + (y-k)^2 = r^2\), \(h\) and \(k\) are directly derived from the equations resulting from completing the square. In the provided problem, after completing the square, the equation was transformed into

• \((x-5)^2 + (y+5)^2 = 50\).

Here, the center of the circle is clearly \((5, -5)\). This is because the expressions \((x-5)\) and \((y+5)\) indicate shifts from the origin (0,0) to the circle's center. Understanding the center helps in plotting the circle on a coordinate plane accurately.

radius of a circle

The radius of a circle is the distance from the center to any point on the circle itself. It's a measure of the circle's size.

In the standard circle equation \((x-h)^2 + (y-k)^2 = r^2\), the term \(r^2\) represents the square of the circle's radius. Once the equation has been rearranged into this standard form, deciphering the radius is straightforward: simply take the square root of \(r^2\).
For instance, in the transformed equation

• \((x-5)^2 + (y+5)^2 = 50\),

we have \(r^2 = 50\).
Hence, \(r = \sqrt{50} = 5\sqrt{2}\), making the radius a valuable piece of information for understanding the geometry and plotting of the circle, since it defines the circle's extent from its center.