A straight line and circle can have two, one or no points of intersection:
If a line and a circle only touch at one point, then the line is a tangent to the circle at that point.
To find out how many times a line and circle meet, we can use substitution.
Problem 1 :
Find the points where the line with equation y = 3x intersects the circle with equation x^{2} + y^{2} = 20 .
Solution :
y = 3x -----(1)
x^{2} + y^{2} = 20 -----(2)
Applying the value of y in (2), we get
x^{2} + (3x)^{2} = 20
x^{2} + 9x^{2}= 20
10x^{2} = 20
x^{2} = 2
x = ±√2
When x = √2, y = 3√2
When x = -√2, y = -3√2
So, the points of intersections are (√2, 3√2) and (-√2, -3√2).
Problem 2 :
Find the points where the line with equation y = 2x + 6 and circle with equation x^{2} + y^{2} + 2x + 2 y − 8 = 0 intersect.
Solution :
y = 2x + 6 ----(1)
x^{2} + y^{2} + 2x + 2 y − 8 = 0 ----(2)
Applying the value of y in (2), we get
x^{2} + (2x + 6)^{2} + 2x + 2(2x + 6) − 8 = 0
x^{2} + 4x^{2} + 24x + 36 + 2x + 4x + 12 - 8 = 0
x^{2} + 4x^{2} + 30x + 40 = 0
5x^{2} + 30x + 40 = 0
x^{2} + 6x + 8 = 0
(x + 2) (x + 4) = 0
x = -2 and x = -4
When x = -2, y = 2(-2) + 6 ==> 2
When x = -4, y = 2(-4) + 6 ==> -2
So, the points of intersections are (-2, 2) and (-4, -2).
Problem 3 :
Find the points of intersection of the line y = 2x + 8 and the circle with equation x^{2} + y^{2} + 4x + 2y – 20 = 0.
Solution :
y = 2x + 8 ----(1)
x^{2} + y^{2} + 4x + 2y − 20 = 0 ----(2)
Applying the value of y in (2), we get
x^{2} + (2x + 8)^{2} + 4x + 2(2x + 8) − 20 = 0
x^{2} + 4x^{2} + 32x + 64 + 4x + 4x + 16 - 20 = 0
x^{2} + 4x^{2} + 40x + 80 - 20 = 0
5x^{2} + 40x + 60 = 0
x^{2} + 8x + 12 = 0
(x + 6) (x + 2) = 0
x = -6 and x = -2
When x = -6, y = 2(-6) + 8 ==> -4
When x = -2, y = 2(-2) + 8 ==> 4
So, the points of intersections are (-6, -4) and (-2, 4).
Problem 4 :
Find the points of intersection of the circle
x^{2} + y^{2} – 2x – 4y + 1 = 0
and the line
x + y = 1.
Solution :
x^{2} + y^{2} – 2x – 4y + 1 = 0 ---(1)
x + y = 1 ----(2)
y = 1 - x
x^{2} + (1 - x)^{2} - 2x - 4(1 - x) + 1 = 0
x^{2} + 1 + x^{2 }- 2x - 2x - 4 + 4x + 1 = 0
2x^{2} - 4x + 4x - 2 = 0
2x^{2} - 2 = 0
(x^{2} - 1) = 0
(x + 1)(x - 1) = 0
x = 1 and x = -1
When x = 1, y = 1 - 1 ==> 0
When x = -1, y = 1 - (-1) ==> 2
So, the points of intersections are (1, 0) and (-1, 2).
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM