Question

A tangent is drawn to a circle with centre (-1, -3). The point of contact with the circle is (2, -1). How do you find the equation of the tangent.?

Answer 1

`y = -3/2x + 2`

Explanation:

It's given that the centre is at `(-1, 3)`, and that the point of tangency (which would be on the circumference of the circle) is at `(2, -1)`. Therefore, the radius can be found using the distance formula.

`d = sqrt((x_2 - x_1)^2 + (y_2 -y_1)^2)`

`d = sqrt((-1 - 2)^2 + (-3 - (-1))^2)`

`d = sqrt(9 + 4)`

`d = sqrt(13)`

Therefore, the radius measures `sqrt(13)` units. We know that the equation of a circle is given by

`(x - a)^2 + (y - b)^2 = r^2`, where `(a,b)` is the centre and `r` is the radius. Hence the equation is

`(x + 1)^2 + (y + 3)^2 = (sqrt(13))^2`

`(x+ 1)^2 + (y + 3)^2 = 13`

If we expand this, we can get

`x^2 + 2x + 1 + y^2 + 6y + 9 = 13`

`x^2 + y^2 + 2x + 6y - 3 = 0`

Differentiating with respect to `y`, we get:

`2x + 2y(dy/dx) + 2 + 6(dy/dx) = 0`

`2y(dy/dx) + 6(dy/dx) = -2x - 2`

`dy/dx(2y + 6) = -2x - 2`

`dy/dx= (-2x- 2)/(2y + 6)`

`dy/dx = (-x - 1)/(y + 3)`

Since the derivative will represent the slope of the tangent, we can simply plug in the point of tangency.

`dy/dx = (-2 - 1)/(-1 + 3) = -3/2`

Now using the point-slope form of a line, we can get:

`y - y_1 = m(x - x_1)`

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

`y + 1 = -3/2x + 3`

`y = -3/2x + 2`

This is the equation of the tangent. Attached is a graph including the circle and the tangent.

Hopefully this helps!