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Chapter 1: Problem 66

Find the center and radius of the circle whose equation is given. $$x^{2}+y^{2}+10 x-75=0$$

Short Answer

Expert verified
Answer: The center of the circle (or the imaginary circle in this case) is at point (-5, 0), and the radius is imaginary with a value of \(-\sqrt{50}\).

Step by step solution

01

Complete the square for x terms

First, we want to complete the square for the \(x\) terms, which are \(x^2 + 10x\). To do this, take half of the coefficient of the \(x\) term (which is 10) and square it. In this case, it is \((\frac{10}{2})^2 = 25\). Add 25 to both sides of the equation to obtain: $$ x^2 + 10x + 25 + y^2 = 25 - 75 $$
02

Complete the square for y terms

In our given equation, there is no term for y so the square is already complete for the y terms, which we can rewrite as \(y^2 + 0y\). Hence, our equation now looks like: $$ x^2 + 10x + 25 + y^2 = - 50 $$
03

Rewrite in the standard form of a circle

Now we rewrite the equation as \((x - a)^2 + (y - b)^2 = r^2\). We have completed the square, so express the x and y terms as perfect squares: $$ (x + 5)^2 + (y + 0)^2 = -50 $$ Notice that the right side of the equation is negative. It shouldn't be negative for a standard circle. This equation represents an imaginary circle, as circles cannot have a negative radius squared. Let's write the final standard form using variables for the center and radius, even though it represents an imaginary circle: $$ (x - (-5))^2 + (y - 0)^2 = (-\sqrt{50})^2 $$
04

Identify the center and radius of the circle

From the standard form of the equation, we can identify the center and radius as follows: - Center \((a, b) = (-5, 0)\) - Radius \(r = -\sqrt{50}\) (imaginary radius) Although the equation represents an imaginary circle, its center is still a valid point in the coordinate system. Remember that circles cannot have a negative radius, so this equation does not represent a real circle.

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