Question

part 2: circle equations (unit 1)

1. what is the radius of this circle based on the following equation:

$(x - 2)^2 + (y + 1)^2 = 4$

1. what is the center of this circle based on the following equation:

$(x - 2)^2 + (y + 1)^2 = 4$

Explanation:

Question 4

Step1: Recall circle equation formula

The standard form of a circle's equation is \((x - h)^2 + (y - k)^2 = r^2\), where \((h,k)\) is the center and \(r\) is the radius.

Step2: Compare with given equation

Given equation: \((x - 2)^2 + (y + 1)^2 = 4\). We can rewrite \(4\) as \(2^2\), so comparing with \((x - h)^2 + (y - k)^2 = r^2\), we have \(r^2 = 2^2\), so \(r = 2\) (radius is non - negative).

Step1: Recall circle equation formula

The standard form of a circle's equation is \((x - h)^2 + (y - k)^2 = r^2\), where \((h,k)\) is the center and \(r\) is the radius.

Step2: Compare with given equation

Given equation: \((x - 2)^2 + (y + 1)^2 = 4\). We can rewrite \(y + 1\) as \(y-(- 1)\). Comparing with \((x - h)^2 + (y - k)^2 = r^2\), we get \(h = 2\) and \(k=-1\).

Answer:

The radius of the circle is 2.

Question 5