Circle Challenges 1
Write the equation in standard form: (x-h)^2+(y-k)^2=r^2
Center (−16, −5) and the circumference is 22pi
(x+16)^2+(y+5)^2=121
Write the equation in standard form: (x-h)^2+(y-k)^2=r^2
Center lies on the y-axis tangent to y = -2 and y = -17
x^2+(y-9.5)^2=56.25
Sketch the graph (accurately), find the vertex, and find the equation of the parabola:
Directrix y= −4, Focus (2, -2)
Vertex: (2, -3)
Equation:
(x-2)^2=4(y+3) or y=1/4(x-2)^2-3
Given the focus and directrix, how can you find the vertex of the parabola?
The vertex has the same x-coordinate as the focus and is the midpoint between the focus and the directrix
Write the equation in standard form: (x-h)^2+(y-k)^2=r^2
Center (13, −27) and the area is 196pi
(x-13)^2+(y+27)^2=196
Write the equation in standard form: (x-h)^2+(y-k)^2=r^2
Three points on the circle are (−8, 5), (3, −6), (14, 5)
(x-11)^2+(y-5)^2=121
The points (0, 5) and (0, -5) are the endpoints of the diameter of a circle. The point (3, y) is on the circle. What is a value for y?
4 or -4
Sketch the graph (accurately), find the vertex, and find the equation of the parabola:
Directrix y= 2, Focus (-1, 0)
Vertex: (-1, 1)
Equation:
(x+1)^2=-4(y-1) or y=-1/4(x+1)^2+1
Given the focus and directrix, how can you tell if the parabola opens up or down?
If the focus is above the directrix, the parabola opens up. If the focus is below the directrix, the parabola opens down.
Write the equation in standard form: (x-h)^2+(y-k)^2=r^2
Diameter measures 15 units and the center is at the intersection of y=x+7 and y=2x-5
(x-12)^2+(y-19)^2=225/4
What is the equation of the circle with center (-13, -16) and containing the point (-10, -16) on the circle?
(x+13)^2+(y+16)^2=9
I know three points on the circle are (-7, 6), (9, 6), and (-4, 13). I think that the equation of the circle is (x −1)^2 + (y − 6)^2 = 64. Is this the correct equation for the circle?
No, the point (-4, 13) does not work with this equation
Sketch the graph (accurately), find the vertex, and find the equation of the parabola:
Directrix y= 3, Focus (1, 7)
Vertex: (1, 5)
Equation:
(x-1)^2=8(y-5) or y=1/8(x-1)^2+5
How do you see the distance between the focus and the vertex (or the vertex and the directrix) showing up in the equations that you have written?
In the equations: (x-h)^2=4p(y-k) or y=1/(4p)(x-h)^2+k the distance between the vertex and the focus is p.
Write the equation in standard form: (x-h)^2+(y-k)^2=r^2
Lies in quadrant 2 tangent to x = −12, y=0 and x = −4
(x+8)^2+(y-4)^2=16
Find the equation of a circle with center in the first quadrant and is tangent to the lines x = 8, y = 3, and x = 14.
(x-11)^2+(y-6)^2=9
Is the point (5, 1) inside, outside, or on the circle x^2−6x+y^2+8y=24
Inside
Sketch the graph (accurately), find the vertex, and find the equation of the parabola:
Directrix y= 3, Focus (2, -1)
Vertex: (2, 1)
Equation:
(x-2)^2=-8(y-1) or y=-1/8(x-2)^2+1
Describe a pattern for writing the equation of a parabola given the focus and directrix.
Answers vary. In general, you set the distance to a point (x, y) to the focus equal to the distance from the point (x, y) to the directrix.
Write the equation in standard form: (x-h)^2+(y-k)^2=r^2
Center (-14, 9) point on circle (1, 11)
(x+14)^2+(y-9)^2=229
The points (4, -1) and (-6, 7) are the endpoints of the diameter of a circle. What is the equation of the circle?
(x+1)^2+(y-3)^2=41
The circle defined by ( x− 1)^2 + (y + 4)^2 = 16 is translated 5 units to the left and 2 units down. Write the equation of the resulting circle.
(x+4)^2+(y+6)^2=16
What is the focus and directrix for y=x^2?
(0, 1/4)
Are parabolas defined geometrically (with a focus and a directrix) quadratic functions? Use an example.
Yes, they have a constant second difference. (If the directrix is parallel to the y-axis)