Find the points where the line y = 2x – 1 meets the circle of radius 3 centered at ( 1, 2).?

Question

Find the points where the line y = 2x – 1 meets the circle of radius 3 centered at ( 1, 2).?

Answer 1

equation of circle:

(x-1)² + (y-2)² = 3²

Substitute 2x-1 for y:

(x-1)² + ((2x-1)-2)² = 3²

solve for x:

(x-1)² + (2x-3)² = 3²

(x²-2x+1) + (4x²-12x+9) = 9

5x² -14x + 1 = 0

x₁ = 2.7266

x₂ = 0.0734

y₁ = 2x₁-1 = 4.4533

y₂ = 2x₂-1 = -0.8533

The line and circle meet at (2.7266, 4.4533) and (0.0734, -0.8533)

http://www.flickr.com/photos/dwread/2807...

Answer 2

would advice you to draw the line, then plot the point.

the circle will either: not cross the line

                              just touch the line, ie it is tangental

                              or it will cut in to places

if you work out the equation for the arch then you can prove it mathematically

or best to draw a line with a length of 3 from your centre point

good luck

Answer 3

Let the points be (h,k)

so, this points lies on the line y = 2x - 1

=> k = 2h - 1 .............. (i)

Now, The equation of the circle is:

(x - 1)^2 + (y - 2)^2 = 3^2

so (h , k) will satisfy this eqn. also.

=> (h - 1)^2 + (k - 2)^2 = 3^2

=> (h - 1)^2 + (2h - 1 - 2)^2 = 9

=> h^2 - 2h + 1 + 4h^2 + 9 - 12h = 9

=> 5h^2 - 14h + 10 - 9 = 0

=> 5h^2 - 14h + 1 = 0

=> h = {14 +/- sqrt[196 - 4.5.1]}/2.5

        = {14 +/- sqrt[196 - 20]}/10

        = {14 +/- sqrt(176)}/10

Now find h, then find k from (i)