Answer:
(a) center: (1, -2); radius √10
(b) distance from (4, -1) to center is √10, so is on the circle
(c) work shown below
(d) y = -1/3x +5/3
Step-by-step explanation:
It is well you should have difficulty, as point B is not defined. If we assume it is (0, -5), then we can proceed as follows.
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(a) The center of the circle will be the midpoint of AB, so will be the average of their coordinates.
C = (A+B)/2 = ((2, 1) +(0, -5))/2 = (2, -4)/2
C = (1, -2) . . . . the center of the circle
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The radius of the circle is the distance between the center and either end of the diameter. (It can also be figured as half the diameter.)
Here, we choose ...
r = ║A-C║ = ║(2, 1) -(1, -2)║ = ║(1, 3)║
r = √(1² +3²)
r = √10 . . . . the radius of the circle
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(b) There are a couple of ways to show (4, -1) is a point on the circle. One of them is to show that its distance from the center is the same as the radius. That distance is ...
║(4, -1) -(1, -2)║ = ║(3, 1)║ = √(3² +1²) = √10 . . . . same as the radius
(4, -1) is on the circle
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(c) The standard form equation of a circle with center (h, k) and radius r is ...
(x -h)² +(y -k)² = r²
To get to the desired general form, we need to eliminate parentheses and subtract r². Filling in the values (h, k) = (1, -2) and r = √10, we have ...
(x -1)² +(y +2)² = 10
x² -2x +1 + y² +4y +4 = 10 . . . . . eliminate parentheses
x² +y² -2x +4y -5 = 0 . . . . . . . . subtract 10 and put in appropriate order
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(d) We can write the equation of the tangent at point A starting with a point-slope form of the equation for the line, then rearranging it to slope-intercept form.
You know the tangent to a circle is perpendicular to the radius at the point of tangency. You know from the work in part (a) that the segment from C to A has a rise of 3 and a run of 1, so a slope of 3/1 = 3. The tangent is perpendicular, so its slope is the negative reciprocal of that, or -1/3.
Now we know the slope of the tangent and the point of tangency (A), so we can write the equation as ...
y -1 = (-1/3)(x -2)
Adding 1 and eliminating parentheses, we get ...
y = -1/3x +2/3 +1
y = -1/3x +5/3
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Comment on notation above
We have used the notation ║(a, b)║ to refer to the length of a segment with components (a, b). It is fully equivalent to √(a²+b²) in this context.
