CAL-PREP ANALYTIC GEOMETRY REVIEW2 |

**E/ CIRCLES:**

**Definition:** A CIRCLE is the path of all points that are exacly

radius (*r*) distance away from the center point (*h*, *k*).

Therefore, the equation of a circle must state algebraically that the distance from any point (*x*, *y*) on the circumference to the center point (*h*, *k*) is equal to the radius ( *r* ).

But, since the distance formula uses that big old ugly square root, we actually state that

the** square** of the distance from any point (*x*, *y*) on the circumference

to the center point (*h, k*) is equal to **square** of the radius (*r*).

The **STANDARD FORM** of the equation of the circle with center (*h*, *k*) and radius *r* is:

**(***x*** – ***h***) ^{2} + (**

If *h* = *k* = 0 making the center at the origin, the equation is: ** x** ²

__ex__: Write the equation of the circle center (**–** 2, 3) with radius 4.

The square of the distance from any point (*x*, *y*) on the circumference of the circle to the center point (*h*, *k*) is [*x* **–** (**–** 2)]² + [*y* **–** (+3)]² or [*x* + 2]² + [*y* **–** 3]².

This must equal the square of the radius or 16.

So, the equation of the circle center (–2,3) with radius 4 is **[ x + 2]^{ }² + [y – 3]² = 16** .

__ex__: The equation of the circle center (0, 0) with radius 5 is:

**[***x*** – 0]² + [***y*** – 0]² = 25 **or** ***x***² + ***y***² = 25** .

**FINDING THE CENTER AND THE RADIUS**:

The **GENERAL FORM** of the equation of a circle is:

*x*^{2} + *y*^{2} – 2*hx*** – 2***ky*** + c = 0, ** where *c*** = ***h*^{2} + *k*^{2} – *r*** ^{2}** .

It is the expansion of **(***x*** – ***h***) ^{2} + (**

So, given the expanded equation, we can find the center coordinates by taking **– (1/2) the coefficients of x and y**. And we can determine the radius knowing

*c*** = ***h*^{2} + *k*^{2} – *r*** ^{2 }** to get

__ex__: Find the center and radius of the circle with equation *x*^{2} + *y*^{2} – 6*x*** + 4***y*** – 12 = 0**

__solution__: (–1/2)(– 6) = 3, (–1/2)(4) = – 2 so the center is (3, – 2).

** c** = –12 so

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**Note:** For this approach to work, the coefficients of *x*² and *y*² must be 1 (one).

If we have, *ax² + by² + ...*, we first divide the equation through by *a*.

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__EXERCISE 5__:

1)Write the equation of the circle:

a) center (0,0) radius 3 | b) center (1,4) radius 2. |

c) center (–2, –5) radius 1. | d) center (3, –10) radius 4. |

e) center (0,0); A(– 1, 6) , B(3, 15) radius = AB |
f) center = midpoint of (–3, –3) & (–7, 1), radius = 5 |

g) center (2, – 4) tangent to the x–axis. |
h) center (2, – 4) tangent to the y–axis. |

i) radius = 4, tangent to both axes (4 answers) | j) radius = 4, tangent to positive x–axis & negative y–axis. |

k) radius = 4, tangent to the positive x–axis and the line x = 1. | |

l) With the line joining (5, – 7) and (1, – 3) as diameter. |

2) Find the center and radius of these circles:

a) x^{2} + y^{2} = 36 |
b) (x – 6)^{2} + (y + 1)^{2} = 16 |

c) x^{2} + y^{2} –12x + 2y + 21 = 0 |
d) x^{2} + y^{2} –2x + 8y + 13 = 0 |

e) 2x^{2} + 2y^{2} + 2x – 6y = 28 |
f) x^{2} + y^{2} + 14x + 24 = 0 |

g) x^{2} + y^{2} – 6y = 0 |
h) x^{2} + y^{2} – 4x + 10y = –29 |

i) x^{2} + y^{2} + 4x + 6y – 23 = 0 |

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__EXERCISE 5__ Solutions

1)

a) x^{2} + y^{2} = 9 |
b) (x – 1)^{2} + (y – 4)^{2} = 4 |

c) (x + 2)^{2} + (y + 5)^{2} = 1 |
d) (x – 3)^{2} + (y + 10)^{2} = 16 |

e) x^{2} + y^{2} = 97 |
f) (x + 5)^{2} + (y + 1)^{2} = 25 |

g) (x – 2)^{2} + (y + 4)^{2} = 1 |
h) (x – 2)^{2} + (y + 4)^{2} = 4 |

i) (x ! 4)^{2} + (y ! 4)^{2} = 16 |
j) (x – 4)^{2} + (y + 4)^{2} = 16 |

k) (x – 5)^{2} + (y ! 4)^{2} = 16 |
l) (x – 3)^{2} + (y + 5)^{2} = 8 |

2)

a) center (0, 0) radius 6 | b) center (6, –1) radius 4 |

c) center (6, –1) radius 4 | d) center (1, –4) radius 2 |

e) center (–1/2, 3/2) radius = | f) center (–7, 0) radius 5 |

g) center (0, 3) radius = 3 | h) center (2, –5) radius 0 |

i) center (–2, –3) radius = 6 |

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