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Analytic Geometry EOCT
UNIT 6: MODELING GEOMETRY
REVIEW EXAMPLES
1) What is the equation of the circle with a center at (4, 5) and a radius of 2?
Solution:
Use the standard form for the equation for a circle, ( x  h) 2  ( y  k ) 2  r 2 . Substitute the
values into the equation, with h = 4, k = 5, and r = 2.
(x – h)2 + (y – k)2 = r 2 Equation for a circle.
(x – (4))2 + (y – (5))2 = (2)2 Substitute the values in the equation of a circle.
(x – 4)2 + (y – 5)2 = 4
Evaluate.
The equation of a circle with a center at (4, 5) and a radius of 2 is (x – 4)2 + (y – 5)2 = 4.
2) What is the center and radius of the circle given by 8x2 + 8y2 – 16x – 32y + 24 = 0?
Solution:
Write the equation in standard form to identify the center and radius of the circle. First, write
the equation so the x-terms are next to each other and the y-terms are next to each other, both
on the left side of the equation, and the constant term is on the right side of the equation.
8x2 + 8y2 – 16x – 32y + 24 = 0
8x2 + 8y2 – 16x – 32y = –24
8x2 – 16x + 8y2 – 32y = –24
x2 – 2x + y2 – 4y = –3
(x2 – 2x) + (y2 – 4y) = –3
Original equation.
Subtract 24 from both sides.
Commutative Property of Addition.
Divide both sides by 8.
Associative Property of Addition.
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Analytic Geometry EOCT
UNIT 6: MODELING GEOMETRY
Next, to write the equation in standard form, complete the square for the x-terms and the
2
 b 
y-terms. Find   for the x- and y-terms.
 2a 
2
2
2
2
 b   2 
2
x-term:    
  (1)  1
 2a   2(1) 
 b   4 
2
y-term:    
  (2)  4
 2a   2(1) 
(x2 – 2x + 1) + (y2 – 4y + 4) = –3 + 1 + 4
(x – 1)2 + (y – 2)2 = 2
Add 1 and 4 to each side of the equation.
Write the trinomials as squares of binomials.
This equation for the circle is written in standard form, where h = 1, k = 2, and r 2 = 2. The
center of the circle is (1, 2) and the radius is 2 .
3) Write an equation for the parabola with a focus at (6, 4) and a directrix of y = 2.
Solution:
The standard form of an equation of a parabola with a horizontal directrix is:
1
( x  h)2 , where (h, k) is the vertex, (h, k + p) is the focus, x = h is the axis of
yk 
4p
symmetry, and y = k – p is the equation of the directrix.
Because the directrix is below the focus, the parabola opens upward.
Based on the x-coordinate of the focus, h = 6.
Based on the y-coordinate of the focus, k + p = 4.
Based on the equation for the directrix, k – p = 2.
Solve a system of equations to find k and p.
k  p  4

k  p  2
2k = 6
k=3
3+p=4
p=1
Add the equations.
Division Property of Equality.
Substitute the value of k in the equation to find p.
Subtraction Property of Equality.
Substitute 6 for h, 3 for k, and 1 for p in the standard form of a parabola, y  k 
An equation for the parabola with a focus at (6, 4) and a directrix of y = 2 is
1
y  3  ( x  6) 2 .
4
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1
( x  h) 2 .
4p
Analytic Geometry EOCT
UNIT 6: MODELING GEOMETRY
REVIEW EXAMPLES
1) A circle is centered at the origin and has a radius of 5 units. A line with a slope of 2 passes
through the origin and intersects the circle in two places. Where does the line intersect the
circle?
Solution:
First write an equation for the circle: x2 + y2 = 5.
Then write an equation for the line: y = 2x.
Solve the system of equations algebraically.
x2  y 2  5
x 2  (2 x)2  5
y  2x
x2  4 x2  5
5x2  5
x2  1
x  1
To find the y-coordinate for the solutions, you can substitute the values for x into either
equation. Substitute 1 and –1 into the linear equation and solve for y.
y  2x
y  2x
y  2(1)
y2
y  2(1)
y  2
The solutions are (1, 2) and (–1, –2), so the line intersects the circle at (1, 2) and (–1, –2).
For the system of equations given above, you can also graph the equations on a coordinate
plane. The line appears to intersect the circle at (–1, –2) and (1, 2). Check these solutions by
substituting them into both equations.
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Copyright © 2013 by the Georgia Department of Education • All Rights Reserved
Analytic Geometry EOCT
UNIT 6: MODELING GEOMETRY
2) Given the system of equations:
x2 + y2 = 25
y = 3x
a. Sketch a graph of the system to determine the number of solutions.
b. Solve the system algebraically to find the solutions.
Solution:
a. The first equation is a circle with a radius of 5 and a center at (0, 0). To sketch the circle,
plot points 5 units from the center: (5, 0), (–5, 0), (0, 5), and (0, –5). Connect the points to
form a circle. The second equation is a line with a slope of 3 that goes through point
(0, 0). To sketch the line, plot the point (0, 0). Use the slope 3 to identify another point,
(1, 3), and connect the points to form a line.
The line intersects the circle at two points, so there are 2 solutions.
b. Solve the system algebraically.
x2 + y2 = 25
y = 3x
Substitute 3x from the second equation for y in the first equation. Then solve for x.
x 2  (3 x) 2  25
x 2  9 x 2  25
10 x 2  25
x 2  2.5
x   2.5
x  1.58
Substitute 3x for y.
Power of a product.
Add.
Divide both sides of the equation by 10.
Take the square root of both sides of the equation.
Evaluate and round.
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Copyright © 2013 by the Georgia Department of Education • All Rights Reserved
Analytic Geometry EOCT
UNIT 6: MODELING GEOMETRY
To find the y-coordinates of the solutions, substitute the values for x   2.5 and
x  2.5 into either equation and solve for y.
y  3x
y  3x
y  3 2.5
y  4.74
y  3 2.5
y  4.74
The solutions are approximately (–1.58, –4.74) and (1.58, 4.74).
EOCT Practice Items
1) A circle is centered at the origin and has a radius of 3 units. A horizontal line passes
through the point (0, 3). In how many places does the line intersect the circle?
A. 0
B. 1
C. 2
D. infinitely many
[Key: B]
2) A circle is centered at the origin and has a radius of 10 units. A line has a slope of –3
and passes through the origin. At which points does the line intersect the circle?
A. (–3, 1) and (3, –1)
B. (–1, 3) and (1, –3)
C. (1, 3) and (–1, –3)
D. (3, 1) and (–3, –1)
[Key: B]
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Copyright © 2013 by the Georgia Department of Education • All Rights Reserved
Analytic Geometry EOCT
UNIT 6: MODELING GEOMETRY
REVIEW EXAMPLES
1) Quadrilateral ABCD has vertices A(–1, 3), B(3, 5), C(4, 3), and D(0, 1). Is ABCD a rectangle?
Explain how you know.
Solution:
First determine whether or not the figure is a parallelogram. If the figure is a parallelogram,
then the diagonals bisect each other. If the diagonals bisect each other, then the midpoints of
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Analytic Geometry EOCT
UNIT 6: MODELING GEOMETRY
the diagonals are the same point. Use the midpoint formula to determine the midpoints for
each diagonal.
 x  x y  y   1  4 3  3   3 6 
Midpoint AC :  1 2 , 1 2   
,
   ,   (1.5, 3)
2   2
2  2 2
 2
 x  x y  y   3  0 5 1   3 6 
Midpoint BD :  1 2 , 1 2   
,
   ,   (1.5, 3)
2   2
2  2 2
 2
The diagonals have the same midpoint; therefore, the diagonals bisect each other and the
figure is a parallelogram.
A parallelogram with congruent diagonals is a rectangle. Determine whether or not the
diagonals are congruent.
Use the distance formula to find the length of the diagonals:
AC = (4 - (-1)) 2 + (3 - 3) 2 = (5) 2 + (0) 2 = 25 + 0 = 25 = 5
BD = (0 - 3)2 + (1- 5)2 = (-3)2 + (-4)2 = 9 +16 = 25 = 5
The diagonals are congruent because they have the same length.
The figure is a parallelogram with congruent diagonals, so the figure is a rectangle.
2) Circle C has a center of (–2, 3) and a radius of 4. Does point (–4, 6) lie on circle C?
Solution:
The distance from any point on the circle to the center of the circle is equal to the radius. Use
the distance formula to find the distance from (–4, 6) to the center (–2, 3). Then see if it is
equal to the radius, 4.
(4  (2))2  (6  3)2
Substitute the coordinates of the points in the distance formula.
(2)2  (3)2
Evaluate within parentheses.
49
13
Evaluate the exponents.
Add.
The distance from (–4, 6) to (–2, 3) is not equal to the radius, so (–4, 6) does not lie on the
circle. (In fact, since 13  4, the distance is less than the radius, so the point lies inside of
the circle.)
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Copyright © 2013 by the Georgia Department of Education • All Rights Reserved
Analytic Geometry EOCT
UNIT 6: MODELING GEOMETRY
3) A parabola has its focus at (3, 4) and its directrix at y = –2. Does the point (–9, 13) lie on the
parabola?
Solution:
Each point on the parabola is equidistant from the focus and the directrix, y = –2. Because the
directrix is a horizontal line, you can find the distance from any point to the directrix using
the y-coordinates. The distance from point (–9, 13) to y = –2 is |13 – (–2)| = 15. If the given
point is on the parabola, the distance between the point and the focus must also be 15.
Use the distance formula d  ( x2  x1 )2  ( y2  y1 )2 to determine the distance from the given
point to the focus. Then see if it is equal to the distance between that point and the directrix.
(9  3)2  (13  4)2
Substitute the coordinates of the points in the distance formula.
(12)2  (9)2
Evaluate within parentheses.
144  81
225
15
Evaluate the exponents.
Add.
Evaluate the square root.
The point (–9, 13) is the same distance from the focus and the directrix, so point (–9, 13) lies
on the parabola.
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