Question

a circle is the set of points equidistant from a fixed point called the center.

1. complete the standard equation of the circle using the information in the diagram.

$(x - \underline{quad})^2 + (y - \underline{quad})^2 = \underline{quad}^2$

1. juanita is asked to write the standard equation of a circle with center $(-2, 3)$ that passes through the point $(-2, 6)$. is juanita’s equation correct? explain.

juanita’s equation: $(x + 2)^2 - (y + 3)^2 = 9$

1. write the standard equation of each circle from the given information.

a center $(2, -4)$; point $(6, -4)$
$(x - \underline{quad})^2 + (y - (\underline{quad}))^2 = \underline{quad}$
$(x - \underline{quad})^2 + (y + (\underline{quad}))^2 = \underline{quad}$
b. center $(0, 2)$; point $(3, -2)$
$(x - \underline{quad})^2 + (y - \underline{quad})^2 = \underline{quad}$
$x^2 + (y - \underline{quad})^2 = \underline{quad}$
c
center: $\underline{quad}$; radius: $\underline{quad}$
equation: $\underline{quad}$
d.
center: $\underline{quad}$; radius: $\underline{quad}$
equation: $\underline{quad}$

Explanation:

Problem 1

Step1: Recall the standard circle equation

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. From the diagram, the center is \((h,k)\) and the radius is \(r\). So we substitute \(h\) for the first blank, \(k\) for the second blank, and \(r\) for the third blank.

Step2: Fill in the blanks

Using the standard form, the equation is \((x - h)^2 + (y - k)^2 = r^2\). So the blanks are filled with \(h\), \(k\), and \(r\) respectively.

Step1: Recall the standard circle equation

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. The center here is \((-2,3)\), so \(h=-2\) and \(k = 3\).

Step2: Calculate the radius

The circle passes through \((-2,6)\). The radius \(r\) is the distance between the center \((-2,3)\) and the point \((-2,6)\). Using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), since \(x_1=x_2=-2\), the distance is \(|6 - 3|=3\), so \(r = 3\) and \(r^2=9\).

Step3: Analyze Juanita's equation

Juanita's equation is \((x + 2)^2-(y + 3)^2 = 9\). But the standard form has a plus sign between the two squared terms, not a minus. Also, the \(y\)-term should be \((y - 3)\) (since \(k = 3\), so \(y - k=y - 3\), and \((x - (-2))=(x + 2)\) is correct for the \(x\)-term). So Juanita's equation is incorrect because the standard form uses a plus sign between the two squared terms, and the \(y\)-term should be \((y - 3)\) not \((y + 3)\).

Step1: Identify the center and radius

The center is \((2,-4)\), so \(h = 2\) and \(k=-4\). The point on the circle is \((6,-4)\). The radius \(r\) is the distance between \((2,-4)\) and \((6,-4)\). Since the \(y\)-coordinates are the same, the distance is \(|6 - 2| = 4\), so \(r = 4\) and \(r^2=16\).

Step2: Write the equation in two forms

First form: Using \(k=-4\), the equation is \((x - 2)^2+(y - (-4))^2=16\) which simplifies to \((x - 2)^2+(y + 4)^2=16\). Second form (filling the blanks as per the question): For \((x - \_\_)^2+(y - (\_\_\_))^2=\_\_\_\), we have \(h = 2\), \(k=-4\) (so the second blank is \(-4\)), and \(r^2 = 16\). For \((x - \_\_)^2+(y + (\_\_\_))^2=\_\_\_\), since \(y - (-4)=y + 4\), the second blank is \(4\) (because \(y+4=y-(-4)\)), and the rest are \(2\) and \(16\).

Answer:

\((x - \boldsymbol{h})^2 + (y - \boldsymbol{k})^2 = \boldsymbol{r}^2\)

Problem 2