HOW TO SHOW A LINE DOES NOT INTERSECT A CIRCLE

Problem 1 :

Show that the line y = 2x + 1 does not intersect the circle with equation x² + y² - 2x + 4y + 1 = 0

Solution :

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

x² + y² - 2x + 4y + 1 = 0 -------(2)

Applying the value of y in (2), we get

x² + (2x + 1)² - 2x + 4(2x + 1) + 1 = 0

x² + 4x² + 4x + 1 - 2x + 8x + 4 + 1 = 0

5x² + 10x + 6 = 0

Finding the nature of this quadratic equation, we get

a = 5, b = 10 and c = 6

Discriminant = b² - 4ac

= 10² - 4(5) (6)

= 100 - 120

= -20 < 0

The roots are not real, so the circle and line does not touch each other.

Problem 2 :

Show that the line y = 2x + 1 is a tangent to the circle

x² + y² - 6x - 4y + 8 = 0

Solution :

y = 2x + 1  ----(1)

x² + y² - 6x - 4y + 8 = 0 ----(2)

Applying the value of y in (2), we get

x² + (2x + 1)² - 6x - 4(2x + 1) + 8 = 0

x² + 4x² + 4x + 1 - 6x - 8x - 4 + 8 = 0

5x² - 10x + 5 = 0

x² - 2x + 1 = 0

Discriminant = b² - 4ac

a = 1, b = -2 and c = 1

= (-2)² - 4(1) (1)

= 4 - 4

= 0

So, the line is intersecting the circle.

Problem 3 :

By solving the equations as a system, find the points common to the line with equation 𝑥 − 𝑦 = 6 and the circle with equation

x² + y² = 26

Graph the line and the circle to show these points (5, -1) and (1, -5).

Solution :

𝑥 − 𝑦 = 6

x = 6 + y 

x² + y² = 26

(6 + y)² + y² = 26

6² + 2(6)y + y² + y² = 26

y² + y²  + 36 + 12y - 26 = 0

2y² + 12y + 10 = 0

y² + 6y + 5 = 0

(y + 1)(y + 5) = 0

y = -1 and y = -5

Applying y = -1, we get x = 6 - 1 ==> 5

Applying y = -5, we get x = 6 - 5 ==> 1

So, the points of intersections are (5, -1) and (1, 5)

Problem 4 :

Graph the line 5x + 6y = 12 and the circle given by x² + y² = 1. Find all solutions to the system of equations.

Solution :

Equation of line is 5x + 6y = 12

Equation of circle is that x² + y² = 1

6y = 12 - 5x

y = 2 - (5x/6)

Applying the value of y in the equation of circle, we get 

x² + (2 - (5x/6))² = 1

x² + 4 - (10x/3) + (5x/6)² = 1

x² + 4 - (10x/3) + (25x²/36) = 1

36x² + 25x² + 144 - 120x = 36

61x² - 120x + 144 - 36 = 0

61x² - 120x + 108 = 0

Finding the nature of this quadratic equation, we get

a = 61, b = -120 and c = 108

Discriminant = b² - 4ac

= (-120)² - 4(61) (107)

= 14400 - 26108 < 0

So, the roots are not real. then the circle and line will not intersect each other.

Problem 5 :

Graph the line given by 3x + 4y = 25 and the circle given by x² + y² = 25. Find all solutions to the system of equations. Verify your results both algebraically and graphically.

Solution :

Equation of line is that 3x + 4y = 25

Equation of circle is that x² + y² = 25

4y = 25 - 3x

y = (25 - 3x)/4

x² + (25 - 3x)² /16 = 25

16x² + 625 - 150x + 9x² = 400

25x² - 150x + 625 - 400 = 0

25x² - 150x + 225 = 0

x² - 6x + 9 = 0

(x - 3)(x - 3) = 0

x = 3 and x = 3

Applying the value of x, we get

y = (25 - 3(3))/4

= (25 - 9)/4

= 16/4

y = 4

So, the point of intersection of the line and the circle is (3, 4).