In each item, determine the standard equation of the circle satisfying the given conditions.
5. Center (-1,-2), P(5,2)
6. center (5,-6), tangent to y-axis
7. center (5.-6), tangent to the x-axis
8. center (5,-6), tangent to line y = -3
9. center (5,-6), tangent to line x = -7
10. It has a diameter with endpoints A(-1,4) and B(4,2)
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Answer:
To find the standard equation of a circle we use the formula:
(x - h)^2 + (y - k)^2 = r^2
where (h k) represents the center of the circle and r represents the radius. Here's how we can find the equation for each given condition:
5. Center (-1 -2 P(5 2):
The center coordinates are (h k) = (-1 -2) and a point on the circle is P(5 2). We can use the distance formula to find the radius:
r^2 = (x2 - x1)^2 + (y2 - y1)^2
r^2 = (5 - (-1))^2 + (2 - (-2))^2
r^2 = (6)^2 + (4)^2
r^2 = 36 + 16
r^2 = 52
Therefore the equation is:
(x + 1)^2 + (y + 2)^2 = 52
6. Center (5 -6 tangent to y-axis:
Since the circle is tangent to the y-axis the distance from the center to the y-axis will represent the radius. The center coordinates are (h k) = (5 -6) and the y-axis intersects the x-axis at x = 0.
r = |x - h|
r = |0 - 5|
r = 5
Therefore the equation is:
(x - 5)^2 + (y + 6)^2 = 5^2
(x - 5)^2 + (y + 6)^2 = 25
7. Center (5 -6 tangent to the x-axis:
Similarly the distance from the center to the x-axis represents the radius. The center coordinates are (h k) = (5 -6) and the x-axis intersects the y-axis at y = 0.
r = |y - k|
r = |0 - (-6)|
r = 6
Therefore the equation is:
(x - 5)^2 + (y + 6)^2 = 6^2
(x - 5)^2 + (y + 6)^2 = 36
8. Center (5 -6 tangent to line y = -3:
The distance from the center to the line y = -3 represents the radius. The center coordinates are (h k) = (5 -6) and the y-coordinate of the point on the line is -3.
r = |y - k|
r = |-3 - (-6)|
r = 3
Therefore the equation is:
(x - 5)^2 + (y + 6)^2 = 3^2
(x - 5)^2 + (y + 6)^2 = 9
9. Center (5 -6 tangent to line x = -7:
Similarly the distance from the center to the line x = -7 represents the radius. The center coordinates are (h k) = (5 -6) and the x-coordinate of the point on the line is -7.
r = |x - h|
r = |-7 - 5|
r = 12
Therefore the equation is:
(x - 5)^2 + (y + 6)^2 = 12^2
(x - 5)^2 + (y + 6)^2 = 144
10. It has a diameter with endpoints A(-1 4) and B(4 2):
The center of a circle lies halfway between the endpoints of its diameter. We can find the center coordinates using the midpoint formula:
h = (x1 + x2) / 2
k = (y1 + y2) / 2
h = (-1 + 4) / 2 = 3/2
k = (4 + 2) / 2 = 3
Therefore the center coordinates are (h k) = (3/2 3). The radius is half the length of the diameter which can be found using the distance formula:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
d = sqrt((4 - (-1))^2 + (2 - 4)^2)
d = sqrt((5)^2 + (-2)^2)
d = sqrt(25 + 4)
d = sqrt(29)
r = d/2 = sqrt(29)/2
Therefore the equation is:
(x - 3/2)^2 + (y - 3)^2 = (sqrt(29)/2)^2
(x - 3/2)^2 + (y - 3)^2 = 29/4