Let P(x_1, y_1 ) and Q(x_2, y_2 ) be any two points in a plane, as shown in the figure. Hence, the distance ‘d’ between the points P and Q is d = √(〖(x_2-x_1)〗^2+〖〖(y〗_2-y_1)〗^2 ). This is called the distance formula.

**Find the distance between two points A(4, 3) and B(8, 6)**.

**Solution:** Compare these points with (x_1, y_1 ), (x_2, y_2 )

x_1 =4, y_1 =3, x_2=8, y_2 =6

d = √(〖(x_2-x_1)〗^2+〖〖(y〗_2-y_1)〗^2 )

= √(〖(8-4)〗^2+〖(6-3)〗^2 )

= √(4^2+3^2 )

= √(16+9)

=√25

d = 5 Units**Find the distance between two points A(3, 1) and B(6, 4).**

**Solution:** Compare these points with (x_1 , y_1 ), (x_2 , y_2 )

x_1 =3, y_1 =1, x_2=6, y_2 =4

d = √(〖(x_2-x_1)〗^2+〖〖(y〗_2-y_1)〗^2 )

= √(〖(6-3)〗^2+〖(4-1)〗^2 )

= √(3^2+3^2 )

= √(9+9)

=√18

= √(9×2)

d =3 √2 Units ≈4.24 Units

**Center**: The point in the plane that all points of the circle are equidistant to.**Radius:** The line that represents the distance from any given point on the circle to the center.

Since the radius can be expressed as the distance between the center and all points around it, we can use the distance formula to make an equation for the circle.

Write an equation of a circle with a radius r and a center at the origin.

Let (x, y) represent any point on a circle with the center at the origin and radius r.

By the Pythagorean Theorem, x^2+y^2=r^2.

This is the equation of a circle with a radius r and a center at the origin.

**Example 1****a. **Write the equation of the circle shown.

**Solution:**

The radius is 3, and the center is at the origin.

x^2+y^2=r^2 Equation of circle.

x^2+y^2=3^2 Substitute.

x^2+y^2=9 Simplify.

The equation of the circle is x^2+y^2=9^2**b.** Write the standard equation of the circle shown on the graph.

`You can see that the center of the circle is at the origin (0, 0), and the radius is 2 units. Thus,`

x^2+y^2=r^2 Standard equation of a circle going through the origin

x^2+y^2=2^2 Substitute.

x^2+y^2=4 Simplify.

**c.** Write the standard equation of the circle shown on the graph.

You can see that the center of the circle is at the origin (0,0), and the radius is 20 units. Thus,

x^2+y^2=r^2 Standard equation of a circle going through the origin

〖 x〗^2+y^2=20^2 Substitute.

x^2+y^2=400 Simplify.

**Standard Equation of a Circle**

**CIRCLES CENTERED at (h, k)**

You can write the equation of any circle if you know its radius and the coordinates of its center.

Suppose a circle has a radius r and center (h, k).

Let (x, y) be a point on the circle.

The distance between (x, y) and (h, k) is r, so by the Distance Formula,

√(〖(x-h)〗^2+〖(y-k)〗^2 )=r

Square both sides to find the standard equation of a circle.**KEY CONCEPT**

The standard equation of a circle with center (h, k) and radius r is:

〖(x-h)〗^2+〖(y-h)〗^2=r^2**Example 2**

Write the standard equation of a circle with center and radius.

Center (0, -9) and radius 4.2

Center (2, 0), radius 2.5.

Center (-2, 5), radius 7

Write the standard equation of the circle shown on the graph.

Center (0, -9) and radius 4.2.

**Solution:**

〖(x-h)〗^2+〖(y-h)〗^2=r^2 Standard equation of a circle

〖(x-0)〗^2+〖(y-(-9))〗^2=〖4.2〗^2 Substitute.

x^2+〖(y+9)〗^2=17.64 Simplify.

`Write the standard equation of a circle with center (2, 0) and radius 2.5.`

**Solution:**

h=2, k=0 and r=2.5

〖(x-h)〗^2+〖(y-h)〗^2=r^2 Standard equation of a circle.

〖(x-2)〗^2+〖(y-0)〗^2=〖2.5〗^2 Substitute.

〖(x-2)〗^2+y^2=6.25 Simplify.

Write the standard equation of a circle with a center (-2, 5) and radius of 7.**Solution:**

h=-2, k=5 and r=7

〖(x-h)〗^2+〖(y-h)〗^2=r^2 Standard equation of a circle.

〖(x-(-2))〗^2+〖(y-5)〗^2=7^2 Substitute.

〖(x+2)〗^2+〖(y-5)〗^2=49 Simplify.

You can see that the center of the circle is at the point (2, 3), and the radius is two units. Thus,

〖(x-a)〗^2+〖(y-b)〗^2=r^2 Standard equation of a circle.

〖(x-2)〗^2+〖(y-3)〗^2=2^2 Substitute.

〖(x-2)〗^2+〖(y-3)〗^2=4 Simplify.**Example 3**

The point (-5, 6) is on a circle with a center (-1, 3). Next, write the standard equation of the circle.

`The point (3, 4) is on a circle whose center is (1, 4). `

Write the standard equation of the circle.

**Example 3:**

The point (-1, 2) is on a circle whose center is (2, 6). Write the standard equation of the circle**Solutions:**

a. Step 1:

To write the standard equation, you need to know the values of h, k, and r.

To find r, find the distance between the center and the point (-5, 6) on the circle.

r = √(〖(x-h)〗^2+〖(y-k)〗^2 ) Distance formula

r = √(〖[-5-(-1)]〗^2+〖(6-3)〗^2 )

= √(〖(-5+1)〗^2+〖(3)〗^2 ) Simplify

= √(〖(-4)〗^2+3^2 )

= √(16+9)

=√25

∴r = 5

`The point (3, 4) is on a circle whose center is (1, 4). Write the standard equation of the circle.`

**Solution:**

To write the standard equation, you need to know the values of h, k, and r.

To find r, find the distance between the center and the point (3, 4) on the circle.

h = 1, k =4, x=3 and y= 4

r = √(〖(x-h)〗^2+〖(y-k)〗^2 ) Distance formula

r = √(〖(3-1)〗^2+〖(4-4)〗^2 )

= √(2^2+0^2 ) Simplify

= √4

r =2

Substitute (h, k) = (1, 4) and r = 2 into the standard equation of a circle.

〖(x-h)〗^2+〖(y-k)〗^2=r^2 Standard equation of a circle.

〖(x-1)〗^2+〖(y-4)〗^2=2^2 Substitute.

〖(x-1)〗^2+〖(y-4)〗^2=4 Simplify.

The standard equation of the circle is 〖(x-1)〗^2+〖(y-4)〗^2=4

The point (-1, 2) is on a circle whose center is (2, 6). Write the standard equation of the circle.**Solution:**

To write the standard equation, you need to know the values of h, k, and r.

To find r, find the distance between the center and the point (-1, 2) on the circle.

h =1, k =4, x=3 and y= 4

r = √(〖(-1-2)〗^2+〖(2-6)〗^2 ) Distance formula

r = √(〖(-3)〗^2+〖(-4)〗^2 )

= √(9+16) Simplify

= √25

r = 5 Units

Substitute (h, k) = (2, 6) and r = 5 into the standard equation of a circle.

〖(x-h)〗^2+〖(y-k)〗^2=r^2 Standard equation of a circle.

〖(x-2)〗^2+〖(y-6)〗^2=5^2 Substitute.

〖(x-2)〗^2+〖(y-6)〗^2=25 Simplify.

The standard equation of the circle is 〖(x-2)〗^2+〖(y-6)〗^2=25**Example 4:**

The equation of a circle is 〖(x-4)〗^2+〖(y+2)〗^2=36. Graph the circle.

The standard equation of a circle is 〖(x+2)〗^2+〖(y-2)〗^2=4. Graph the circle.

The standard equation of a circle is x^2+〖(y-1)〗^2=16. Graph the circle.

The standard equation of a circle is 〖(x+2)〗^2+〖(y-3)〗^2=25. Graph the circle.

The equation of a circle is 〖(x-4)〗^2+〖(y+2)〗^2=36. Graph the circle.

S**olution:**

Rewrite the equation to find the center and radius.

〖(x-4)〗^2+〖(y+2)〗^2=36

〖(x-4)〗^2+[〖(y-(-2)]〗^2=6^2

The center is (4, -2), and the radius is 6.

Use a compass to graph the circle.

`The equation of a circle is 〖(x+2)〗^2+〖(y-2)〗^2=4. Graph the circle.`

**Solution:**

Rewrite the equation to find the center and radius.

〖(x+2)〗^2+〖(y-2)〗^2=4

〖(x-(-2)〗^2+〖(y-2)〗^2=2^2

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

Use a compass to graph the circle.

C. The standard equation of a circle is x^2+〖(y-1)〗^2=16. Graph the circle.**Solution:**

Rewrite the equation to find the center and radius.

x^2+〖(y-1)〗^2=16

〖(x-0)〗^2+〖(y-1)〗^2=4^2

The center is (0, 1), and the radius is 4.

Use a compass to graph the circle.

d. The standard equation of a circle is 〖(x+2)〗^2+〖(y-3)〗^2=25. Graph the circle.**Solution:**

Rewrite the equation to find the center and radius.

〖[x-(-2)]〗^2+〖(y-3)〗^2=25

〖[x-(-2)]〗^2+〖(y-3)〗^2=5^2

The center is (-2, 3), and the radius is 5.

Use a compass to graph the circle.

Use graphs of circles**Example 5**:

Earthquakes

The epicenter of an earthquake is the point on Earth’s surface directly above the earthquake’s origin. A seismograph can be used to determine the distance to the epicenter of an earthquake. Seismographs are needed in three different places to locate an earthquake’s epicenter. Use the seismograph readings from locations A, B, and C to find the epicenter of an earthquake.

The epicenter is 7 miles away from A(-2, 2.5).

The epicenter is 4 miles away from B(4, 6).

The epicenter is 5 miles away from C(3, -2.5).

**Solution:**

The set of all points equidistant from a given point is a circle, so the epicenter is located on each of the following circles.

A with center (-2, 2.5) and radius 7

B with center (4, 6) and radius 4

C with center (3, -2.5) and radius 5

To find the epicenter, graph the circles on a graph where units are measured in miles. Then, find the point of intersection of all three circles.

The epicenter is at about (5, 2).

Check Your Knowledge:

Write the equation of a circle shown below.

```
Write the standard equation of a circle with center (2, -4) and radius 5.
The point (2, 3) is on a circle with center (4, 1). Write the standard equation of the circle.
The standard equation of a circle is 〖(x-4)〗^2+〖(y-1)〗^2=1. Graph the circle.
Determine whether the given equation defines a circle. If the equation defines a circle, rewrite the equation in standard form.
```

x^2-8x+16+y^2+2y+4=25

**Solution:**

The radius is 4, and the center is at the origin.

x^2+y^2=r^2

x^2+y^2=4^2

x^2+y^2=16

The equation of the circle is x^2+y^2=16**Solution:**

h=2, k=-4 and r=5

Standard equation of a circle

〖(x-h)〗^2+〖(y-k)〗^2=r^2

〖(x-2)〗^2+〖[y-(-4)]〗^2=5^2

〖(x-2)〗^2+〖(y+4)〗^2=25

**Solution:****Step 1: **Point (2, 3) center (4, 1).

Point (x, y) center (h, k).

Find radius

r = √(〖(x-h)〗^2+〖(y-k)〗^2 )

= √(〖(2-4)〗^2+〖(3-1)〗^2 )

= √(〖(-2)〗^2+〖(2)〗^2 )

= √(4+4)

= √8

= 2√2

r≈2.8

**Step 2: **Substitute (h, k) = (4, 1) and r =2.8 into the standard equation of a circle.

is 〖(x-h)〗^2+〖(y-k)〗^2=r^2

〖(x-4)〗^2+〖(y-1)〗^2=〖2.8〗^2

〖(x-4)〗^2+〖(y-1)〗^2=7.84

**Solution:**

Rewrite the equation to find the center and radius.

〖(x-4)〗^2+〖(y-1)〗^2=1

〖(x-4)〗^2+〖(y-1)〗^2=1^2

〖(x-h)〗^2+〖(y-k)〗^2=r^2

The center is (4, 1), and the radius is 1. Use a compass to graph the circle.

Determine whether the given equation defines a circle. If the equation defines a circle, rewrite the equation in standard form.

x^2-8x+16+y^2+2y+4=25**Solution:**

x^2-8x+16+y^2+2y+4=25

x^2-8x+4^2+y^2+2y+2^2=5^2

x^2-2.x.4+4^2+y^2+2.y.1+2^2=5^2

〖(x-4)〗^2+〖(y-2)〗^2=5^2 By using algebraic identity formulas

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