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## 2 Ways to Graph a Circle

Circles are simple to work with in pre-calculus. A circle has one center, one radius, and a whole lot of points, but you follow slightly different steps, depending on whether you are graphing a circle centered at the origin or away from the origin.

The first thing you need to know in order to graph the equation of a circle is where on a plane the center is located. The equation of a circle appears as (*x – h*)^{2} + (*y – v*)^{2} = *r*^{2}. This is called the *center-radius* *form* (or standard form) because it gives you both pieces of information at the same time.

The *h* and *v* represent the coordinates of the center of the circle being at the point (*h, v*), and *r* represents the radius. Specifically, *h *represents the horizontal displacement — how far to the left or to the right of the *y-*axis the center of the circle is. The variable *v* represents the vertical displacement — how far above or below the *x-*axis the center falls.

From the center, you can count from the center *r* units (the radius) horizontally in both directions and vertically in both directions to get four different points, all equidistant from the center. Connect these four points with the best curve that you can sketch to get the graph of the circle.

## Graphing circles centered at the origin

The simplest circle to graph is one whose center is at the origin (0, 0). Because both *h* and *v* are zero, they can disappear and you can simplify the standard circle equation to look like *x*^{2} + *y*^{2} = *r*^{2}. For instance, to graph the circle *x*^{2} + *y*^{2} = 16, follow these steps:

- Realize that the circle is centered at the origin (no
*h*and*v*) and place this point there. - Calculate the radius by solving for
*r.*Set*r*^{2}= 16. In this case, you get*r*= 4. - Plot the radius points on the coordinate plane.You count out 4 in every direction from the center (0, 0): left, right, up, and down.
- Connect the dots to graph the circle using a smooth, round curve.

The figure shows this circle on the plane.

## Graphing circles centered away from the origin

Although graphing circles at the origin is easiest, very few graphs are as straightforward and simple as those. In pre-calculus, you work with transforming graphs of all different shapes and sizes. Fortunately, these graphs all follow the same pattern for horizontal and vertical shifts, so you don’t have to remember many rules.

**Don’t forget that the coordinates of the center of the circle are of the opposite signs of the h and v from inside the parentheses in the equation. Because the h and v are inside the grouping symbols, this means that the shift happens opposite from what you would think.**

For example, follow these steps to graph the equation (*x* – 3)^{2} + (*y* + 1)^{2} = 25:

- Locate the center of the circle from the equation (
*h, v*).(*x*– 3)^{2}means that the*x-*coordinate of the center is positive 3.(*y*+ 1)^{2}means that the*y-*coordinate of the center is negative 1.Place the center of the circle at (3, –1). - Calculate the radius by solving for
*r.*Set*r*^{2}= 25 and square root both sides to get*r*= 5. - Plot the radius points on the coordinate plane.Count 5 units up, down, left, and right from the center at (3, –1). This step gives you points at (8, –1), (–2, –1), (3, –6), and (3, 4).
- Connect the dots to the graph of the circle with a round, smooth curve.

The figure shows a visual representation of this circle.

## How to Graph a Circle

The first thing you need to know in order to graph the equation of a circle is where on a plane the center is located. The equation of a circle appears as

This is called the *center-radius*form (or standard form) because it gives you both pieces of information at the same time. The *h* and *v* represent the center of the circle at point (*h*, *v*), and *r* names the radius. Specifically, *h* represents the horizontal displacement — how far to the left or to the right the center of the circle falls from the *y*-axis. The variable *v* represents the vertical displacement — how far above or below the center falls from the *x*-axis.

You can count from the center *r* units (the radius) horizontally in both directions and vertically in both directions. This will give you four different points, all equidistant from the center. Connect these four points with the best curve that you can sketch to get the graph of the circle.

## Center at the origin

The simplest circle to graph has its center at the origin (0, 0). Because both *h* and *v* are zero, they can disappear and you can simplify the standard circle equation to look like

- Realize that the circle is centered at the origin (no
*h*and*v*) and place this point there. - Calculate the radius by solving for
*r*.Set*r*-squared = 16. In this case, you get*r*= 4. - Plot the radius points on the coordinate plane.You count out 4 in every direction from the center (0, 0): left, right, up, and down.
- Connect the dots to graph the circle using a smooth, round curve.

## Center away from the origin

Graphing a circle anywhere on the coordinate plane is pretty easy when its equation appears in center-radius form. All you do is plot the center of the circle at (*h*, *k*), and then count out from the center *r* units in the four directions (up, down, left, right). Then, connect those four points with a nice, round circle.

Unfortunately, while it is much easier to graph circles at the origin, very few are as straightforward and simple as those. In pre-calc, you work with transforming graphs of all different shapes and sizes (this is nothing new to you, right?). Fortunately, these graphs all follow the same pattern for horizontal and vertical shifts, so you don’t have to remember many rules.

**Don’t forget to switch the sign of the h and v from inside the parentheses in the equation. This is necessary because the h and v are inside the grouping symbols, which means that the shift happens opposite from what you would think.**

For example, to graph the equation

## Circle Equations

## Circle on a Graph

Let us put a circle of radius 5 on a graph:

And then use Pythagoras:

x^{2} + y^{2} = 5^{2}

There are an infinite number of those points, here are some examples:

In all cases a point on the circle follows the rule x^{2} + y^{2} = radius^{2}

We can use that idea to find a missing value

**Example: x value of 2, and a radius of 5**

^{2}+ y

^{2}= r

^{2}

^{2}+ y

^{2}= 5

^{2}

^{2}= 5

^{2}− 2

^{2}

^{2}− 2

^{2})

**±4.58…**

*(The ± means there are two possible values: one with + the other with −)*

And here are the two points:

## More General Case

Now let us put the center at **(a,b)**

So the circle is **all the points (x,y)** that are **“r”** away from the center **(a,b)**.

Now lets work out where the points are (using a right-angled triangle and Pythagoras):

It is the same idea as before, but we need to subtract **a** and **b**:

(x−a)^{2} + (y−b)^{2} = r^{2}

And that is the **“Standard Form”** for the equation of a circle!

It shows all the important information at a glance: the center **(a,b)** and the radius **r**.

### Example: A circle with center at (3,4) and a radius of 6:

Start with:

(x−a)^{2} + (y−b)^{2} = r^{2}

Put in (a,b) and r:

(x−3)^{2} + (y−4)^{2} = 6^{2}

We can then use our algebra skills to simplify and rearrange that equation, depending on what we need it for.

## “General Form”

But you may see a circle equation and **not know it**!

Because it may not be in the neat “Standard Form” above.

As an example, let us put some values to a, b and r and then expand it

^{2}+ (y−b)

^{2}= r

^{2}

^{2}+ (y−2)

^{2}= 3

^{2}

^{2}− 2x + 1 + y

^{2}− 4y + 4 = 9

^{2}+ y

^{2}− 2x − 4y + 1 + 4 − 9 = 0

And we end up with this:

x^{2} + y^{2} − 2x − 4y − 4 = 0

It is a circle equation, but “in disguise”!

So when you see something like that think *“hmm … that might be a circle!”*

In fact we can write it in **“General Form”** by putting constants instead of the numbers:

x^{2} + y^{2} + Ax + By + C = 0

*Note: General Form always has x ^{2} + y^{2} for the first two terms*.

## Going From General Form to Standard Form

Now imagine we have an equation in **General Form**:

x^{2} + y^{2} + Ax + By + C = 0

How can we get it into **Standard Form** like this?

(x−a)^{2} + (y−b)^{2} = r^{2}

The answer is to Complete the Square (read about that) twice … once for **x** and once for **y**:

### Example: x^{2} + y^{2} − 2x − 4y − 4 = 0

^{2}+ y

^{2}− 2x − 4y − 4 = 0

**x**s and

**y**s together:(x

^{2}− 2x) + (y

^{2}− 4y) − 4 = 0

^{2}− 2x) + (y

^{2}− 4y) = 4

Now complete the square for **x** (take half of the −2, square it, and add to both sides):

(x^{2} − 2x + (−1)^{2}) + (y^{2} − 4y) = 4 + (−1)^{2}

And complete the square for **y** (take half of the −4, square it, and add to both sides):

(x^{2} − 2x + (−1)^{2}) + (y^{2} − 4y + (−2)^{2}) = 4 + (−1)^{2} + (−2)^{2}

Tidy up:

^{2}− 2x + 1) + (y

^{2}− 4y + 4) = 9

^{2}+ (y − 2)

^{2}= 3

^{2}

And we have it in Standard Form!

(Note: this used the a=1, b=2, r=3 example from before, so we got it right!)

## Unit Circle

If we place the circle center at (0,0) and set the radius to 1 we get:

## How to Plot a Circle by Hand

1. Plot the center **(a,b)**

2. Plot 4 points “radius” away from the center in the up, down, left and right direction

3. Sketch it in!

### Example: Plot (x−4)^{2} + (y−2)^{2} = 25

The formula for a circle is (x−a)^{2} + (y−b)^{2} = r^{2}

So the center is at (4,2)

And **r ^{2}** is

**25**, so the radius is √25 = 5

So we can plot:

- The Center: (4,2)
- Up: (4,2+5) = (4,7)
- Down: (4,2−5) = (4,−3)
- Left: (4−5,2) = (−1,2)
- Right: (4+5,2) = (9,2)

Now, just sketch in the circle the best we can!

## How to Plot a Circle on the Computer

We need to rearrange the formula so we get “y=”.

We should end up with two equations (top and bottom of circle) that can then be plotted.

### Example: Plot (x−4)^{2} + (y−2)^{2} = 25

So the center is at (4,2), and the radius is √25 = 5

Rearrange to get “y=”:

^{2}+ (y−2)

^{2}= 25

^{2}to the right:(y−2)

^{2}= 25 − (x−4)

^{2}

^{2}]

*(notice the ± “plus/minus” …*

there can be two square roots!)

there can be two square roots!)

^{2}]

So when we plot these two equations we should have a circle:

- y = 2 + √[25 − (x−4)
^{2}] - y = 2 − √[25 − (x−4)
^{2}]

Try plotting those functions on the Function Grapher.

It is also possible to use the Equation Grapher to do it all in one go.

## ACT Math : How to find the equation of a circle

### Example Question #1 : How To Find The Equation Of A Circle

In the standard (x,y) coordinate plane, what is the radius and the center of the circle (x−5)^{2}+(y+4)^{2}=27 ?

### Example Question #2 : How To Find The Equation Of A Circle

In the standard (x,y) coordinate plane, what is the area of the circle x^{2}+y^{2}=169 ?

**Possible Answers:**

13π

12π

169π

26π

28,561π

**Correct answer:**

169π

Explanation:

The general equation of a circle is x^{2}+y^{2}=r^{2}^{ }.

According to the question, r^{2}=169. Thus, r=13.

The general equation for the area of a circle is A=πr^{2}.

When we plug in 13 for r, we get our area to equal 169π.

### Example Question #1 : How To Find The Equation Of A Circle

A circle in the standard coordinate plane is tangent to the x-axis at (3,0) and tangent to the y-axis at (0,3). What is the equation of the circle?

**Possible Answers:**

The formula for the equation of a circle is (x – h)^{2}+ (y – k)^{2 }= r^{2},

where (h, k) represents the coordinates of the center of the circle, and r represents the radius of the circle.

If a circle is tangent to the x-axis at (3,0), this means it touches the x-axis at that point. If a circle is tangent to the y-axis at (0,3), this means it touches the y-axis at that point. Given these two points, we can determine the center and the radius of the circle. The center of the circle must be equidistant from any of the points on the circumference. This means that both (0,3) and (3,0) are the same distance from the center. If we draw these points on a coodinate plane, it becomes apparent that the center of the circle must be (3,3). This point is exactly three units from each of the given points, indicating that the radius of the circle is 3.

When we input this information into the formula for a circle, we get (x – 3)^{2 }+ (y – 3)^{2 }= 9.

### Example Question #4 : How To Find The Equation Of A Circle

Find the equation of the circle with center coordinates of (2,4) and a radius of 3.

**Possible Answers:**

### Example Question #5 : How To Find The Equation Of A Circle

On the xy plane, what is the area of a circle with the following equation:

(x+2)^{2}+(y−3)^{2}=64

**Possible Answers:**

13π

8π

50π

16π

64π

**Correct answer:**

64π

Explanation:

The standard form equation of a circle is (x−h)^{2}+(y−h)^{2}=r^{2},

where (h,k) is the center of the circle and r is equal to the radius. Thus, since we have the circle’s standard form equation already given to us, we can ignore h and k, since all we need is r^{2}.

The area of circle is equal to πr^{2}, which is equal to 64π.

A circle has a center at (5,5) and a radius of 2. If the format of the equation for the circle is (x-A)^{2}+(y-B)^{2}=C, what is C?

**Possible Answers:**

3

4

1

5

2

**Correct answer:**

4

Explanation:

The circle has a center at (5,5) and a radius of 2. Therefore, the equation is (x-5)^{2}+(y-5)^{2}=2^{2}, or (x-5)^{2}+(y-5)^{2}=4.

### Example Question #1 : How To Find The Equation Of A Circle

If the center of a circle is at (0,4) and the diameter of the circle is 6, what is the equation of that circle?

**Possible Answers:**

x^{2} + y^{2 }= 9

(x-4)^{2} + y^{2} = 9

x^{2} + (y-4)^{2} = 36

x^{2} + (y-4)^{2} = 9

(x-4)^{2} + y^{2} = 36

**Correct answer:**

x^{2} + (y-4)^{2} = 9

Explanation:

The formula for the equation of a circle is:

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

Where (h,k) is the center of the circle.

h = 0 and k = 4

and diameter = 6 therefore radius = 3

(x-0)^{ 2} + (y-4)^{2} = 3^{2}

x^{2} + (y-4)^{2} = 9

### Example Question #8 : How To Find The Equation Of A Circle

Circle A is given by the equation (x – 4)^{2} + (y + 3)^{2} = 29. Circle A is shifted up five units and left by six units. Then, its radius is doubled. What is the new equation for circle A?

**Possible Answers:**

(x – 2)^{2} + (y + 2)^{2} = 58

(x – 10)^{2} + (y + 8)^{2} = 116

(x + 2)^{2} + (y – 2)^{2} = 58

(x – 10)^{2} + (y + 8)^{2} = 58

(x + 2)^{2} + (y – 2)^{2} = 116

**Correct answer:**

(x + 2)^{2} + (y – 2)^{2} = 116

Explanation:

The general equation of a circle is (x – h)^{2} + (y – k)^{2} = r^{2}, where (h, k) represents the location of the circle’s center, and r represents the length of its radius.

Circle A first has the equation of (x – 4)^{2} + (y + 3)^{2} = 29. This means that its center must be located at (4, –3), and its radius is √29.

We are then told that circle A is shifted up five units and then left by six units. This means that the y-coordinate of the center would increase by five, and the x-coordinate of the center would decrease by 6. Thus, the new center would be located at (4 – 6, –3 + 5), or (–2, 2).

We are then told that the radius of circle A is doubled, which means its new radius is 2√29.

Now, that we have circle A’s new center and radius, we can write its general equation using (x – h)^{2} + (y – k)^{2} = r^{2}.

(x – (–2))^{2} + (y – 2)^{2} = (2√29)^{2} = 2^{2}(√29)^{2} = 4(29) = 116.

(x + 2)^{2} + (y – 2)^{2} = 116.

The answer is (x + 2)^{2} + (y – 2)^{2} = 116.

### Example Question #9 : How To Find The Equation Of A Circle

Which of the following equations describes all the points (x, y) in a coordinate plane that are five units away from the point (–3, 6)?

**Possible Answers:**

(x – 3)^{2} + (y + 6)^{2} = 5

(x + 3)^{2} + (y – 6)^{2} = 25

y + 6 = 5 – (x – 3)^{2}

(x – 3)^{2} + (y + 6)^{2} = 25

(x – 3)^{2} – (y + 6)^{2} = 25

**Correct answer:**

(x + 3)^{2} + (y – 6)^{2} = 25

**Explanation:**

We are trying to find an equation for all of the points that are the same distance (5 units) from (–3, 6). The locus of all points equidistant from a single point is a circle. In other words, we need to find an equation of a circle. The center of the circle will be (–3, 6), and the radius, which is the distance from (–3,6), will be 5.

The standard form of a circle is given below:

(x – h)^{2} + (y – k)^{2} = r^{2}, where the center is located at (h, k) and r is the length of the radius.

In this case, h will be –3, k will be 6, and r will be 5.

(x – (–3))^{2} + (y – 6)^{2} = 5^{2}

(x + 3)^{2} + (y – 6)^{2} = 25

The answer is (x + 3)^{2} + (y – 6)^{2} = 25.

### Example Question #1 : How To Find The Equation Of A Circle

What is the equation for a circle of radius 12, centered at the intersection of the two lines:

*y*_{1} = 4*x* + 3

and

*y*_{2} = 5*x* + 44?

**Possible Answers:**

(x + 41)^{2} + (y + 161)^{2} = 144

None of the other answers

(x – 22)^{2} + (y – 3)^{2} = 12

(x – 41)^{2} + (y – 161)^{2} = 144

(x – 3)^{2} + (y – 44)^{2} = 144

**Correct answer:**

(x + 41)^{2} + (y + 161)^{2} = 144

Explanation:

To begin, let us determine the point of intersection of these two lines by setting the equations equal to each other:

4*x* + 3 = 5*x* + 44; 3 = *x* + 44; –41 = *x*

To find the *y*-coordinate, substitute into one of the equations. Let’s use *y*_{1}:

*y* = 4 * –41 + 3 = –164 + 3 = –161

The center of our circle is therefore: (–41, –161).

Now, recall that the general form for a circle with center at (*x*_{0}, *y*_{0}) is:

(*x* – *x*_{0})^{2} + (*y* – *y*_{0})^{2} = *r*^{2}

For our data, this means that our equation is:

(*x* + 41)^{2} + (*y* + 161)^{2} = 12^{2} or (*x* + 41)^{2} + (*y* + 161)^{2} = 144

## Example Questions

### Example Question #1 : How To Find The Equation Of A Circle

The diameter of a circle has endpoints at points (2, 10) and (–8, –14). Which of the following points does NOT lie on the circle?

**Possible Answers:**

(–8,–12)

(2, –14)

(9,3)

(–8, 10)

(–15, –7)

**Correct answer:**

(–8,–12)

**Explanation:**

### Example Question #1 : How To Find The Equation Of A Circle

A circle exists entirely in the first quadrant such that it intersects the y-axis at y=6. If the circle intersects the x-axis in at least one point, what is the area of the circle?

**Possible Answers:**

6π

Not enough information

144π

36π

9π

**Correct answer:**

36π

Explanation:

We are given two very important pieces of information. The first is that the circle exists entirely in the first quadrant, the second is that it intersects both the x– and y-axis.

The fact that it is entirely in the first quadrant means that it cannot go past the two axes. For a circle to intersect the x-axis in more than one point, it would necessarily move into another quadrant. Therefore, we can conclude it intersects in exactly one point.

The intersection of the circle with (0,6) must also be tangential, since it can only intersect in one point. We can thus conclude that the circle must have both x– and y– intercepts equal to 6 and have a center of (6,6).

This leaves us with a radius of 6 and an area of:

π∗radius^{2}=π∗6^{2}=36π

### Example Question #3 : How To Find The Equation Of A Circle

We have a square with length 2 sitting in the first quadrant with one corner touching the origin. If the square is inscribed inside a circle, find the equation of the circle.

**Possible Answers:**

If the square is inscribed inside the circle, in means the center of the circle is at (1,1). We need to also find the radius of the circle, which happens to be the length from the corner of the square to it’s center.

r^{2}=1^{2}+1^{2}

Now use the equation of the circle with the center and r.

We get (x−1)^{2}+(y−1)^{2}=2

### Example Question #4 : How To Find The Equation Of A Circle

What is the radius of a circle with the equation x(x−8)+y(y−6)=24?

**Possible Answers:**

7

4

6

3

8

**Correct answer:**

7

### Example Question #5 : How To Find The Equation Of A Circle

A circle has its origin at (0,0). The point (5,7) is on the edge of the circle. What is the radius of the circle?

**Possible Answers:**

### Example Question #1 : How To Find The Equation Of A Circle

The endpoints of a diameter of circle A are located at points (−2,4) and (6,8). What is the area of the circle?

**Possible Answers:**

40π

12π

20π

80π

160π

**Correct answer:**

20π

**Explanation:**

The formula for the area of a circle is given by A =πr^{2 }. The problem gives us the endpoints of the diameter of the circle. Using the distance formula, we can find the length of the diameter. Then, because we know that the radius (r) is half the length of the diameter, we can find the length of r. Finally, we can use the formula A =πr^{2} to find the area.

### Example Question #181 : Coordinate Plane

What is the equation for a circle of radius 9, centered at the intersection of the following two lines?

y1=2x+14

y2=4x+26

**Possible Answers:**

**Explanation:**

To begin, let us determine the point of intersection of these two lines by setting the equations equal to each other:

2x+14=4x+26

14=2x+26

−12=2x

x=−6

To find the *y*-coordinate, substitute into one of the equations. Let’s use y1:

y=2⋅(−6)+14

y=−12+14

y=2

The center of our circle is therefore (−6,2).

Now, recall that the general form for a circle with center at (x0,y0) is

### Example Question #182 : Coordinate Plane

What is the equation of a circle with a center of (4,8) and a diameter of 4?

**Possible Answers:**

Recall that the equation of a circle is defined as:

(x−x0)^{2}+(y−y0)^{2}=r^{2}

Where (x0,y0) is the center of the circle.

Given the data we have, we know that the radius of the circle must be 2 (half the diameter). Thus, we know that the equation of the circle in question must be:

(x−4)^{2}+(y−8)^{2}=4

### Example Question #183 : Coordinate Plane

A circle has a diameter defined by the points (−4,10) and (16,10). What is the equation of this circle?

**Possible Answers:**

Recall that the equation of a circle is defined as:

(x−x0)^{2}+(y−y0)^{2}=r^{2}

Where (x0,y0) is the center of the circle.

So, you must first find the center of the circle in question. This you can do by finding the midpoint of the two points given to us. Since they represent a diameter of the circle, their midpoint must be the center of the circle.

Recall that the midpoint of two points is found by the equation:

### Example Question #184 : Coordinate Plane

Which of the following is the equation of a circle with a center at (s,t) with a radius of 20?

**Possible Answers:**

To begin, recall that the equation of a circle is defined as:

### Example Question #185 : Coordinate Plane

What is the equation of a circle centered around the point (2,4) with a radius of 5?

**Possible Answers:**

### Example Question #186 : Coordinate Plane

What is the equation of a circle centered about the origin with a radius of 7? Simplify all exponential expressions if possible.

**Possible Answers:**

### Example Question #187 : Coordinate Plane

What is the equation of a circle with center (4,2) and radius of r=3?

**Possible Answers:**

### Example Question #188 : Coordinate Plane

Which of the following equations describes a circle centered on the x-axis?

**Possible Answers:**

The basic formula for a circle in the coordinate plane is (x−h)^{2}+(y−k)^{2}=r^{2}, where (h,k) is the center of the circle with radius r.

Since k refers to the y-coordinate of the center, and we know that any point on the x-axis has a y-coordinate of 0, we merely need to look for the equation in which k does not exist.

Note that despite meeting this requirement, (x+4)^{2}=7

### Example Question #189 : Coordinate Plane

Circle G has diameter Q, which intersects the circle at points (2,b) and (2,b−10). Given this information, which of the following is an accurate equation for circle G?

**Possible Answers:**

### Example Question #190 : Coordinate Plane

A circle is centered on point (5,12). The area of the circle is 9π. What is the equation of the circle?

**Possible Answers:**

### Example Question #21 : How To Find The Equation Of A Circle

A circle with a radius of five is centered at the origin. A point on the circumference of the circle has an *x*-coordinate of two and a positive *y*-coordinate. What is the value of the *y*-coordinate?

**Possible Answers:**

Recall that the general form of the equation of a circle centered at the origin is:

*x*^{2} + *y*^{2} = *r*^{2}

We know that the radius of our circle is five. Therefore, we know that the equation for our circle is:

*x*^{2} + *y*^{2} = 5^{2}

*x*^{2} + *y*^{2} = 25

Now, the question asks for the positive *y*-coordinate when* x *= 2. To solve this, simply plug in for *x*:

2^{2} + *y*^{2} = 25

4 + *y*^{2} = 25

*y*^{2} = 21

*y* = ±√(21)

Since our answer will be positive, it must be √(21).

### Example Question #821 : Act Math

Which of the following gives the equation of a circle tangent to the line *y=−2* with its center at (3,2)?

**Possible Answers:**

The equation for a circle is (x – h)^{2} + (y – k)^{2} = r^{2}, where (h,k) is the center of the circle and *r* is the radius.

We can eliminate (x+3) +(y+2) = 4 and (x+3)^{2} + (y+2)^{2} = 64. The first equation does not square the terms in parentheses, and the second refers to a center of (-3,-2) rather than (3,2).

Drawing the line y = -2 and a point at (3,2) for the center of the circle, we see that the only way the line could be tangent to the circle is if it touches the bottom-most part of the circle, directly under the center. The point of intersection will be (3,-2). From this, we can see that the distance from the center to this point of intersection is 4 units between (3,2) and (3,-2). This means the radius of the circle is 4.

Use the center point and the radius in the formula for a circle to find the final answer: (x – 3)^{2} + (y – 2)^{2} = 16

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