FIND POINT OF CONTACT OF TANGENT AND CIRCLE

Problem 1 :

Show that the line y = -3x - 10 is the tangent to the circle

x² + y² - 8x + 4y - 20 = 0

and also find the point of contact.

Solution :

x² + y² - 8x + 4y - 20 = 0 ------(1)

y= -3x - 10 ----(2)

Applying the value of y in (1), we get

x² + (-3x - 10)² - 8x + 4(-3x - 10) - 20 = 0

x² + 9x² + 60x + 100 - 8x - 12x - 40 - 20 = 0

10x² + 40x + 40 = 0

x² + 4x + 4 = 0

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

x = -2 and x = -2

In both case, we have two same values for x. It shows the line touches the curve at one point. So it must be tangent. 

Finding point of contact :

When x = -2,

y = -3(-2) - 10 

y = -16

So, the point of contact is (-2, -16).

Problem 2 :

The circle

x² + y² + 4x - 7y - 8 = 0

cuts the y-axis at two points. Find the coordinates of these points.

Solution :

x² + y² + 4x - 7y - 8 = 0

Since the circle cuts the y-axis at two points, at point of intersection the value of x will be 0.

If x = 0

y² - 7y - 8 = 0

(y - 8)(y + 1) = 0

y = 8 and y = -1

So, the points of contact are (0, 8) and (0, -1).

Problem 3 :

The circle

x² + y² - 2x + 10y - 24 = 0

cuts the x-axis at the points A and B. Find the length of AB.

Solution :

Since the circle cuts the x-axis at two points, at point of intersection the value of y will be 0.

If y = 0

x² - 2x - 24 = 0

(x - 6) (x + 4) = 0

x = 6 and x = -4

So, the points of intersection are A(6, 0) and B(-4, 0).