Question

50pts...
I really don't get this can someone help pls I'd appreciate it a lot​

Answer 1

a) centre: (1,-2)

radius = sqrt(10)

b) distance from the centre = radius

c) x² + y² - 2x + 4y - 5 = 0

d) y = (-1/3)x + (5/3)

Explanation:

Centre is the midpoint of the diameter:

(h,k) = (2+0)/2, (1-5)/2

(h,k) = (1,-2)

radius is ½ times length of the diameter:

½ sqrt[(-5-1)² + (0-2)²]

½ sqrt(40)

½ × 2 sqrt(10)

radius = sqrt(10) units

Find the distance between the centre and (4,-1)

sqrt[(4-1)² + (-1--2)²]

sqrt(9 + 1)

sqrt(10) = radius

Therefore on the circle

Equation:

(x - h)² + (y - k)² = r²

(x - 1)² + (y + 2)² = 10

x² - 2x + 1 + y² + 4y + 4 - 10 = 0

x² + y² - 2x + 4y - 5 = 0

Tangent at (2,1)

Slope would be -ve reciprocal of the slope from A to the centre

m = (-2-1)/(1-2) = 3

Slope of the tangent = - ⅓

y - 1 = -⅓(x - 2)

y = (-⅓)x + ⅔ + 1

y = (-1/3)x + (5/3)

Answer 2

(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

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.