HOW TO CHECK IF THE GIVEN LINE IS TANGENT TO CIRCLE

As we have seen, a line is a tangent if it intersects the circle at only one point.

To show that a line is a tangent to a circle, the equation of the line can be substituted into the equation of the circle, and solved there should only be one solution

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 :

Show that the line with equation

x + y = 4

y = 4 - x  -----(1)

is a tangent to the circle with equation

x² + y² + 6x + 2y − 22 = 0  -----(2)

Solution :

x² + y² + 6x + 2y − 22 = 0

By applying (1) in (2), we get

x² + (4 - x)² + 6x + 2(4 - x) − 22 = 0

x² + 16 - 8x + x² + 6x + 8 - 2x - 22 = 0

2x² - 4x + 2 = 0

x² - 2x + 1 = 0

(x - 1) (x - 1) = 0

x = 1 and x = 1

From this, it is clear the line touches the circle at one point.

When x = 1, y = 4 - 1 ==> 3

So, the point of contact of the line and circle is (1, 3) and the given line is tangent for the circle.

Problem 2 :

Given the line

y = 2x - 1

and the circle 

x² + y² - 4x - 6y + 13 = 0

a) Show that the line is tangent to the circle

b) Find the point of contact.

Solution :

x² + (2x - 1)² - 4x - 6(2x - 1) + 13 = 0

x² + 4x² - 4x + 1 - 4x - 12x + 6 + 13 = 0

5x² - 20x + 20 = 0

x² - 4x + 4 = 0

(x - 2)(x - 2) = 0

x = 2 and x = 2

To find point of contact :

When x = 2

y = 2(2) - 1

y = 3

So, the point of contact is (2, 3).

Problem 3 :

Show that the line y = 3x + 1 is common tangent to the two circles

x² + y² + 8x + 2y +7 = 0

and

x² + y² - 2x + 12y + 27 = 0

Solution :

x² + y² + 8x + 2y + 7 = 0 ----(1)

and

x² + y² - 2x + 12y + 27 = 0 ----(2)

Applying y = 3x + 1 in (1),

x² + (3x + 1)² + 8x + 2(3x + 1) + 7 = 0

x² + 9x² + 6x + 1 + 8x + 6x + 2 + 7 = 0

10x² + 20x + 10 = 0

x² + 2x + 1 = 0

(x + 1) (x + 1) = 0

x = -1 and x = -1

When x = -1, y = 3(-1) + 1 ==> -2

(-1, -2) is the point of contact.

Applying y = 3x + 1 in (2),

x² + (3x + 1)² - 2x + 12(3x + 1) + 27 = 0

x² + 9x² + 6x + 1 - 2x + 36x + 12 + 27 = 0

10x² + 40x + 40 = 0

x² + 4x + 4 = 0

(x + 2)(x + 2) = 0

x = -2 and x = -2

When x = -2, y = 3(-2) + 1 ==> -1

(-2, -1) is the point of contact.

Problem 4 :

Determine whether the line 4y - 3x + 20 = 0 is a tangent to the circle x² + y² - 2x - 4y - 20 = 0

Solution :

x² + y² - 2x - 4y - 20 = 0

x² - 2x + y² - 4y - 20 = 0

(x - 1)² - 1² + (y - 2)² - 2² - 20 = 0

(x - 1)² - 1 + (y - 2)² - 4 - 20 = 0

(x - 1)² + (y - 2)² - 25 = 0

(x - 1)² + (y - 2)² = 25

Center of circle (1, 2) 

Radius f the circle = 5 units.

Distance between the a point and a line = |Ax₁ + By₁ + C| / √A² + B²

= |4(2) - 3(1) + 20|/√4² + 3²

= |8 - 3 + 20| / √(16 + 9)

= 25 / √25

= 25 / 5

= 5

4y - 3x + 20 = 0

slope of the line 4y = 3x - 20

y = (3/4) x - (20/4)

y = (3/4) x - 5 -----(1)

Slope of the line = 3/4

Equation of the line which is perpendicular, then slope = -4/3

(y - y₁) = m(x - x₁)

(y - 2) = (-4/3)(x - 1)

y = (-4x/3) + (4/3) + 2

y = (-4x/3) + (10/3) -----(2)

3x/4 - 5 = (-4x/3) + (10/3)

(9x + 16x)/3 = 10/3 + 5

25x/3 = (10 + 15)/3

25x = 25

x = 1

applying x = 1, we get

y = (3/4) - 5

y = (3-5)/4

= -2/4

= -1/2

So, the point of intersection is (1, -1/2).