Question

Find the center and radius of the circle $$ x^{2}-10 x+y^{2}+4 y=35 $$

Short answer

The center is (5, -2) and the radius is 8.

Step-by-step solution

- Write the equation in standard form

The given equation is \[ x^{2} - 10x + y^{2} + 4y = 35 \] To find the center and radius, rewrite the equation by completing the square.

- Complete the square for x

Take the x-terms: \[ x^2 - 10x \] To complete the square, add and subtract \( \bigg( \frac{-10}{2} \bigg)^2=25 \) inside the equation: \[ x^2 - 10x + 25 - 25 \] This becomes: \[ (x - 5)^2 - 25 \]

- Complete the square for y

Take the y-terms: \[ y^2 + 4y \] To complete the square, add and subtract \( \bigg( \frac{4}{2} \bigg)^2=4 \) inside the equation: \[ y^2 + 4y + 4 - 4 \] This becomes: \[ (y + 2)^2 - 4 \]

- Incorporate the completed squares into the equation

Rewrite the original equation using the completed squares: \[ (x - 5)^2 - 25 + (y + 2)^2 - 4 = 35 \] Next, simplify by isolating the completed squares on one side and constant terms on the other: \[ (x - 5)^2 + (y + 2)^2 = 64 \]

- Identify the center and radius

The equation is now in standard form: \[ (x - h)^2 + (y - k)^2 = r^2 \] where the center is \((h, k)\) and the radius is \(r\). In this case: \[ (x - 5)^2 + (y + 2)^2 = 64 \] Thus, the center is \((5, -2)\) and the radius is \( r = \sqrt{64} = 8 \).

Key concepts

circle equation

Circles have a specific type of equation that represents their geometric properties. Typically, a circle equation is written in the general form of a quadratic equation: \(x^2 + y^2 + Dx + Ey + F = 0\).
Here, \(D\), \(E\), and \(F\) are constants. This form may not be immediately useful if you need to find the center and radius. To make it more practical, we need to convert it into a 'standard form'. Let’s understand this by looking at our given example: \(x^2 - 10x + y^2 + 4y = 35\).
The goal is to rearrange and manipulate this equation algebraically to find the circle's center and radius.

completing the square

Completing the square is a powerful technique used in algebra to transform a quadratic equation into a perfect square trinomial.
This makes it easier to simplify and solve.
Here's a step-by-step guide on how to complete the square:

• Take the term that contains the variable (like \(x\) or \(y\)) and its coefficient.
• Divide the coefficient by 2, then square the result.
• Add and subtract this squared number in the equation to balance it.

In our example:
For the \(x\) terms: \(x^2 - 10x\),
we take \( \frac{ -10 }{ 2 } = -5\) and then square it to get 25. We add and subtract 25 around these terms: \(x^2 - 10x + 25 - 25\), simplifying to: \((x-5)^2 - 25\).
Perform similar steps for the \(y\) terms: \(y^2 + 4y\), where \(\frac{4}{2} = 2\) and squaring gives us 4. This adds and subtracts 4 to form: \(y^2 + 4y + 4 - 4\), simplifying to: \((y+2)^2 - 4\).
This method makes our quadratic expressions into neat square terms.

standard form of a circle

To easily identify a circle's center and radius, we rewrite its equation in the standard form: \((x - h)^2 + (y - k)^2 = r^2\).
Here, \((h, k)\) represents the center of the circle and \(r\) stands for the radius.
In our example from the exercise, after completing the squares, we got: \((x - 5)^2 - 25 + (y + 2)^2 - 4 = 35\).
Simplifying this equation, we obtain: \((x - 5)^2 + (y + 2)^2 = 64\).
This is now in the standard form, with center at \( (5, -2) \) and radius given by \( r = \sqrt{64} = 8 \).
Converting a general quadratic equation into this form is crucial for understanding the circle's geometric attributes.

algebraic manipulation

Algebraic manipulation is the process of rearranging algebraic expressions to achieve a desired form or solution.
It requires understanding various algebraic techniques and properties.
In our example, we use algebraic manipulation through the technique of completing the square. Here's how these manipulations worked in our problem:

• First, we separated the \(x\) and \(y\) terms.
• Then, we 'completed the square' for both \(x\) and \(y\) terms.
• Next, we rewrote the equation using these squares and isolated the constant terms.
• Lastly, we simplified to standard form to easily identify the center and radius.

Mastering algebraic manipulation allows you to transform and simplify equations, which is especially useful in geometry and calculus.
It’s a skill worth practicing!