FIND THE EQUATION OF TANGENT OF CIRCLE AT A POINT

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 :

Shown is the circle

(x - 2)² + (y - 4)² = 25

with the point P on its circumference

a) Write down the coordinates of its center C.

b) Find the gradient of the radius CP.

c) Write down the gradient of the tangent through P.

d) Find the equation of the tangent.

Solution :

(x - 2)² + (y - 4)² = 25

a) Comparing with (x - h)² + (y - k)² = r²

It is clear that (h, k) ==> (2, 4)

b) Slope of CP = (y₂ - y₁) / (x₂ - x₁)

C(2, 4) and P(6, 7)

= (7 - 4) / (6 - 2)

= 3/4

c) The line which passes through the point of tangency will be perpendicular to the line drawn from the center of the circle.

Slope of the tangent line passes through P = -4/3

d) Equation of tangent passes through the point P :

(y - y₁) = m (x - x₁)

(y - 7) = -4/3(x - 6)

3(y - 7) = -4(x - 6)

3y - 21 = -x + 24

x + 3y = 24 + 21

x + 3y = 45

Problem 2 :

For the circle x² + y² + 2x - 4y - 12 = 0

a) Find the coordinates of its center C.

b) Show that the point T(3, 1) lies on its circumference.

c) Find the equation of the tangent to the circle through T.

Solution :

a) x² + y² + 2x - 4y - 12 = 0

x² + 2x + y² - 4y - 12 = 0

x² + 2x(1) + 1² - 1² + y² - 2y(2) + 2² - 2² - 12 = 0

(x + 1)² - 1 + (y - 2)²  - 4 - 12 = 0

(x + 1)² + (y - 2)²  - 17 = 0

(x + 1)² + (y - 2)²  = 17

(x + 1)² + (y - 2)²  = √17²

(x - (-1))² + (y - 2)²  = √17²

Center (-1, 2) and radius = √17

b) To show that the point (3, 1) lies of the circumference of the circle will apply the point in the equation of the circle.

(x + 1)² + (y - 2)²  = 17

(3 + 1)² + (1 - 2)²  = 17

4² + (-1)²  = 17

16 + 1 = 17

17 = 17

So, the point (3, 1) lies on the circle.

c) Equation of tangent :

(y - y₁) = m (x - x₁)

Slope of the line joining the point (-1, 2) and (3, 1)

Slope = (1 - 2) / (3 + 1)

= -1/4

Slope of the line tangent = 4

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

y = 4x - 12 + 1

y = 4x - 11

So, the equation of tangent is y = 4x - 11

Problem 3 :

The line y = 4x + c is the tangent to the circle x² + y² = 17

a) Find the two values of c.

b) For the positive value of c, determine the point of contact of the tangent and the circle.

Solution :

y = 4x + c ------(1)

x² + y² = 17 ------(2)

Applying (1) in (2), we get

x² + (4x + c)² = 17

x² + (4x)² + 2(4x)c + c² = 17

17x² + 8cx + c² - 17 = 0

a = 17, b = 8c and c = c² - 17

b² - 4ac = 0

(8c)² - 4(17) (c² - 17) = 0

64c²  - 68c² + 1156 = 0

 -4c² + 1156 = 0

4c² = 1156

c² = 1156/4

c² = 289

c = -17 and 17

When c = -17

17x² + 8(-17)x + (-17)² - 17 = 0

17x² - 136x + 289 - 17 = 0

17x² - 136x + 272 = 0

x² - 8x + 16 = 0

(x - 4)(x - 4) = 0

x = 4 and x = 4

When c = 17

17x² + 8(17)x + (17)² - 17 = 0

17x² + 136x + 289 - 17 = 0

17x² + 136x + 272 = 0

x² + 8x + 16 = 0

(x + 4)(x + 4) = 0

x = -4 and x = -4

When x = 4 and c = 17

y = 4(4) + 17

= 16 + 17

y = 33

When x = -4 and c = 17

y = 4(-4) + 17

= -16 + 17

y = 1

When x = 4 and c = -17

y = 4(4) - 17

= 16 - 17

y = -1

So, the point of contact is at (1, -4) and (-1, 4).