What is the closest distance between the line and the circle?

Question

There is a line with a formula 7x+4y-140=0
and a circle with a formula x²+y²-8x-4y-5=0 in xy plane.
Please don't just give the number. Also give the solution.

Answer 1

OK, gianlino's answer is both correct (I checked) and simpler than mine.

Thanks to gianlino for finding my mistake.

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My thoughts on an outline for computing this:

1) Find the slope of the line (-7/4) and the center of the circle (4, 2).

2) Find the equation of a line with perpendicular slope (4/7) that passes through the center of the circle:

4/7 = (y - 2)/(x - 4)
4(x - 4) = 7(y - 2)
4x - 16 = 7y - 14
4x - 2 = 7y
y = 4/7 x - 2/7

3) Find the point where that line intersects the other, and the nearer of the two points where it intersects the circle.

Line:

4x - 7y = 2
7x + 4y = 140

16x - 28y = 8
49x + 28y = 980

65x = 988
x = 15.2

4(15.2) - 7y = 2
60.8 - 7y = 2
-7y = -58.8
y = 8.4

Circle:

x² + y² - 8x - 4y - 5 = 0
x² + (4/7x - 2/7)² - 8x - 4(4/7x - 2/7) - 5 = 0
x² + 16/49x² - 16/49x + 4/49 - 8x - 16/7x + 8/7 - 5 = 0
49x² + 16x² - 16x + 4 - 392x - 112x + 56 - 245 = 0
65x² - 520x - 185 = 0
13x² - 104x - 37 = 0

x = (104 + 14√65)/26 = (52 + 7√65)/13
y = 4/7 (52 + 7√65)/13 - 2/7 = (182 + 28√65)/91

4) Find the distance between those two points.

d = √((15.2 - (52 + 7√65)/13)² + (8.4 - (182 + 28√65)/91)²)
d = √(((145.6 - 7√65)/13)² + ((582.4 - 28√65)/91)²)
d = √((24384.36 - 2038.4√65)/169 + (390149.76 - 32614.4√65)/8281)
d = √((1194833.64 - 99881.6√65 + 390149.76 - 32614.4√65)/8281)
d = √((1584983.4 - 132496√65)/8281)
d = √(1584983.4 - 132496√65)/91
d = √516766.497/91
d = 7.8996

I tried to get it in the same form as gianlino's, but no luck.

If you calculate both using a calculator, though, they are equal.

Answer 2

The line equation should be rewritten in the standard format: y = mx + c, where m is the slope and c the intercept.

4y = -7x -140

y = (-7/4)x - 35

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From the circle equation, you can find the centre of the circle as a (x,y) position.

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The shortest distance is measured along a line (a new line) from the centre of the circle, perpendicular to the original line.

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lines are perpendicular to each other if their slope are inverse in values and of opposite sign.

The original line has a slope of -7/4. So your new line will have a slope of +4/7 and can be written:

Y = (4/7)X + C

you can find the value of C by plugging in the values for the centre of the circle (you know that this centre must be on the new line).

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Once you have the complete equation for the new line, you can find two points:

The intersection of the circle with the new line (call it point A), and the intersection of the two perpendicular lines (call it point B.

Once A and B are identified, find the distance between them SQRT( (xA-xB)^2 + (yA-yB)^2 )

Answer 3

The circle can be written as (x-4)^2+(y-2)^2=29.The line is 7x+4y=140.x=20-4y/7.substitute x in the equation of circle and solve for x and then find the value of y.This will give the shortest distance between the two.

Answer 4

This simply is the distance of the center to the line, minus the radius of the circle. So you get with center (4,2)

d= | 7*4+4*2-140|/sqrt{49+16} - radius =104/sqrt{65} - sqrt{4*4+2*2+5} =104/sqrt{65} - 5.

Answer 5

A at once line is technically the shortest distance between 2 factors even nevertheless it rather is not consistently the least complicated direction. Even on the globe, it rather is nonetheless the shortest distance. useful, it must be fairly much impossible each and every so often, yet nonetheless that's.