Its circle?
2017-02-11 11:47 pm
Find the equation of circle with radius 4 units, whose center lies on the line 4x+13y=32 and which touches the line 4x+3y+28=0
回答 (2)
2017-02-12 5:55 am
The lines parallel and distance 4 from 4x+3y+28=0 are 4x+3y+28±4√(4²+3²)=0, which simplified and separated are 4x+3y+8=0 and 4x+3y+48=0.
These lines intersect 4x+13y=32 at (-5,4) and (-18,8) respectively, so the equations of the circles are (x+5)²+(y-4)²=16 and (x+18)²+(y-8)²=16 respectively.
Graph: https://www.desmos.com/calculator/c9llhxafkp
These lines intersect 4x+13y=32 at (-5,4) and (-18,8) respectively, so the equations of the circles are (x+5)²+(y-4)²=16 and (x+18)²+(y-8)²=16 respectively.
Graph: https://www.desmos.com/calculator/c9llhxafkp
2017-02-12 12:46 am
Center a circle on 4x + 13y = 32. The center is (h , k)
(x - h)^2 + (y - k)^2 = r^2 is the formula for the circle
4h + 13k = 32
4h = 32 - 13k
h = 8 - 3.25 * k
r = 4
(x - (8 - 3.25 * k))^2 + (y - k)^2 = 4^2
(x + 3.25 * k - 8)^2 + (y - k)^2 = 16
A point on this circle is (x , y) and it passes through 4x + 3y + 28 = 0 once
4x = -3y - 28
x = (-3/4) * y - 7
(x + 3.25 * k - 8)^2 + (y - k)^2 = 16
((-3/4) * y - 7 + (13/4) * k - 8)^2 + (y - k)^2 = 16
((13/4) * k - (3/4) * y - 15)^2 + (y - k)^2 = 16
(1/16) * (13k - 3y - 60)^2 + (y - k)^2 = 16
(13k - 3y - 60)^2 + 16 * (y - k)^2 = 256
(13k - 3y)^2 - 120 * (13k - 3y) + 3600 + 16 * (y - k)^2 = 256
(3y - 13k)^2 + 120 * (3y - 13k) + 3600 + 16 * (y - k)^2 = 256
9y^2 - 78yk + 169k^2 + 360y - 1560k + 3600 + 16y^2 - 32yk + 16k^2 = 256
9y^2 + 16y^2 - 78yk - 32yk + 169k^2 + 16k^2 + 360y - 1560k + 3600 - 256 = 0
25y^2 - 110yk + 185k^2 + 360y - 1560k + 3600 - 256 = 0
25y^2 + 360y - 110yk + 185k^2 - 1560k + 3344 = 0
25y^2 + (360 - 110k) * y + (185k^2 - 1560k + 3344) = 0
y = (110k - 360 +/- sqrt((360 - 110k)^2 - 4 * 25 * (185k^2 - 1560k + 3344))) / (2 * 25)
y = (110k - 360 +/- sqrt(100 * (36 - 11k)^2 - 100 * (185k^2 - 1560k + 3344))) / 50
y = (110k - 360 +/- 10 * sqrt((11k - 36)^2 - 185k^2 + 1560k - 3344)) / 50
y = (11k - 36 +/- sqrt(121k^2 - 792k + 1296 - 185k^2 + 1560k - 3344)) / 5
y = (11k - 36 +/- sqrt(-64k^2 + 768k - 2048)) / 5
y = (11k - 36 +/- sqrt(-64 * (k^2 - 12k + 32))) / 5
y = (11k - 36 +/- 8 * sqrt(-(k^2 - 12k + 36 - 4))) / 5
y = (11k - 36 +/- 8 * sqrt(4 - (k - 6)^2)) / 5
This relates y (the value on 4x + 3y + 28 = 0) and k (the center of the circle, on 4x + 13y = 32). If there's only one value, then sqrt(4 - (k - 6)^2) will be 0
4 - (k - 6)^2 = 0
4 = (k - 6)^2
-2 , 2 = k - 6
k = 6 - 2 , 6 + 2
k = 4 , 8
WHEW!
4x + 13 * 4 = 32
x + 13 = 8
x = -5
4x + 13 * 8 = 32
x + 13 * 2 = 8
x + 26 = 8
x = -18
(-5 , 4) and (-18 , 8) are your circle centers
(x + 5)^2 + (y - 4)^2 = 16
(x + 18)^2 + (y - 8)^2 = 16
(x - h)^2 + (y - k)^2 = r^2 is the formula for the circle
4h + 13k = 32
4h = 32 - 13k
h = 8 - 3.25 * k
r = 4
(x - (8 - 3.25 * k))^2 + (y - k)^2 = 4^2
(x + 3.25 * k - 8)^2 + (y - k)^2 = 16
A point on this circle is (x , y) and it passes through 4x + 3y + 28 = 0 once
4x = -3y - 28
x = (-3/4) * y - 7
(x + 3.25 * k - 8)^2 + (y - k)^2 = 16
((-3/4) * y - 7 + (13/4) * k - 8)^2 + (y - k)^2 = 16
((13/4) * k - (3/4) * y - 15)^2 + (y - k)^2 = 16
(1/16) * (13k - 3y - 60)^2 + (y - k)^2 = 16
(13k - 3y - 60)^2 + 16 * (y - k)^2 = 256
(13k - 3y)^2 - 120 * (13k - 3y) + 3600 + 16 * (y - k)^2 = 256
(3y - 13k)^2 + 120 * (3y - 13k) + 3600 + 16 * (y - k)^2 = 256
9y^2 - 78yk + 169k^2 + 360y - 1560k + 3600 + 16y^2 - 32yk + 16k^2 = 256
9y^2 + 16y^2 - 78yk - 32yk + 169k^2 + 16k^2 + 360y - 1560k + 3600 - 256 = 0
25y^2 - 110yk + 185k^2 + 360y - 1560k + 3600 - 256 = 0
25y^2 + 360y - 110yk + 185k^2 - 1560k + 3344 = 0
25y^2 + (360 - 110k) * y + (185k^2 - 1560k + 3344) = 0
y = (110k - 360 +/- sqrt((360 - 110k)^2 - 4 * 25 * (185k^2 - 1560k + 3344))) / (2 * 25)
y = (110k - 360 +/- sqrt(100 * (36 - 11k)^2 - 100 * (185k^2 - 1560k + 3344))) / 50
y = (110k - 360 +/- 10 * sqrt((11k - 36)^2 - 185k^2 + 1560k - 3344)) / 50
y = (11k - 36 +/- sqrt(121k^2 - 792k + 1296 - 185k^2 + 1560k - 3344)) / 5
y = (11k - 36 +/- sqrt(-64k^2 + 768k - 2048)) / 5
y = (11k - 36 +/- sqrt(-64 * (k^2 - 12k + 32))) / 5
y = (11k - 36 +/- 8 * sqrt(-(k^2 - 12k + 36 - 4))) / 5
y = (11k - 36 +/- 8 * sqrt(4 - (k - 6)^2)) / 5
This relates y (the value on 4x + 3y + 28 = 0) and k (the center of the circle, on 4x + 13y = 32). If there's only one value, then sqrt(4 - (k - 6)^2) will be 0
4 - (k - 6)^2 = 0
4 = (k - 6)^2
-2 , 2 = k - 6
k = 6 - 2 , 6 + 2
k = 4 , 8
WHEW!
4x + 13 * 4 = 32
x + 13 = 8
x = -5
4x + 13 * 8 = 32
x + 13 * 2 = 8
x + 26 = 8
x = -18
(-5 , 4) and (-18 , 8) are your circle centers
(x + 5)^2 + (y - 4)^2 = 16
(x + 18)^2 + (y - 8)^2 = 16
收錄日期: 2021-04-24 00:13:51
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https://hk.answers.yahoo.com/question/index?qid=20170211154702AAr1nG6