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:

(xh)2 + (yk)2 = r 2

If h = k = 0 making the center at the origin, the equation is: x ² + y ² = r 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 x² + y² = 25 .

FINDING THE CENTER AND THE RADIUS:

The GENERAL FORM of the equation of a circle is:

x2 + y2 – 2hx – 2ky + c = 0, where c = h2 + k2r2 .

It is the expansion of (xh)2 + (yk)2 = r 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 = h2 + k2r2 to get r2 = h2 + k2c.

ex: Find the center and radius of the circle with equation x2 + y2 – 6x + 4y – 12 = 0

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

c = –12 so r2 = 32 + (–2)2 – (–12) = 25. r = 5.

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Note: For this approach to work, the coefficients of x² and y² 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) x2 + y2 = 36 b) (x – 6)2 + (y + 1)2 = 16 c) x2 + y2 –12x + 2y + 21 = 0 d) x2 + y2 –2x + 8y + 13 = 0 e) 2x2 + 2y2 + 2x – 6y = 28 f) x2 + y2 + 14x + 24 = 0 g) x2 + y2 – 6y = 0 h) x2 + y2 – 4x + 10y = –29 i) x2 + y2 + 4x + 6y – 23 = 0

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EXERCISE 5 Solutions

1)

 a) x2 + y2 = 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) x2 + y2 = 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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