Question

Find the angle between the tangents to the circle x² + y² - y -3 =0 at the point where x=1 and find their point of intersection?

Answer 1

Angle between tangents (3 d.p.) is `1.176` radians
The tangents intersect at `(4, 1/2)`

Explanation:

Noting values of `y` at `x = 1` satisfy

`1^2 + y^2 - y - 3 = 0`

implies

`y^2 - y - 2 = 0`

This implies (quadratic formula)

`y = (-(-1) +- sqrt((-1)^2 - 4(1)(-2)))/(2(1))`

`y = (1 +- sqrt(9))/2`

`y = 2 " or " y = -1`

(or, roots may be found by factorising as `(y - 2)(y +1) = 0`)

This implies that tangents through `x = 1`
touch the circle at `(1, 2)` and `(1, -1)`

Noting ("completing the square")

`y^2 - y = (y - 1/2)^2 -1/4`

so that the equation of the circle may be written

`x^2 + y^2 - y - 3 = (x - 0)^2 + (y - 1/2)^2 - 13/4 = 0`

Rearranging,

`(x - 0)^2 + (y - 1/2)^2 = (sqrt(13/4))^2`

which implies the circle has centre `(0, 1/2)` and radius `sqrt(13)/2`

The radius that forms the normal to one of the tangents passes from `(0, 1/2)` to `(1, 2)`

Noting that a line passing through points `(x_1, y_1)` and `(x_2, y_2)` has equation:

`(y - y_1) = (y_2 - y_1)/(x_2 - x_1) (x - x_1)`

The line passing through these points has equation

`(y - 1/2) = (2 - 1/2)/(1 - 0) (x - 0)`

rearranging

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

Noting that the slope of a normal to a line is the negative reciprocal of the slope of the line, the tangent associated with this radius has slope `-2/3`.

The other radius passes from `(0, 1/2)` to `(1, -1)`

The line passing through these points has equation

`(y - 1/2) = (-1 - 1/2)/(1 - 0) (x - 0)`

rearranging

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

Noting that the slope of a normal to a line is the negative reciprocal of the slope of the line, the tangent associated with this radius has slope `2/3`.

Noting that, given a slope `m`, a line passing through the point `(x_1, y_1)` has equation

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

The equation of the first tangent to the circle has equation

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

that is

`y = -2/3 x + 8/3`

The equation of the second tangent to the circle has equation

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

that is

`y = 2/3 x - 5/3`

Noting that the angle between lines with slopes `m_1` and `m_2` has tangent

`(m_1 - m_2)/(1 + m_1 * m_2)`

The angle between the tangents is the arctangent of

`(-2/3 - 2/3)/(1 + (-2/3)(2/3)) = -12/9`

that is, the required angle (as an absolute value) is

`1.176` radians (three decimal places).

The tangents intersect when values of `x` and `y` are equal.

Equal values of `y` imply

`-2/3 x + 8/3 = 2/3 x - 5/3`

Rearranging

`(2/3 + 2/3) x = 8/3 + 5/3`

that is

`4/3 x = 13/3`

implies

`x = 13/4`

This implies (substituting this value of x into the equation for the line corresponding to one of the tangents)

`y = (13/4) (2/3) - 5/3 = 26/12 - 20/12 = 6/12 = 1/2`

That is,

`y = 1/2`.

Thus, the lines intersect at `(4, 1/2)`

I made several arithmetic errors in early versions of this answer. I think I have spotted and corrected them all but remember to check workings.

One reason why I thought that there were errors is that the workings show reflective symmetry in the slopes of the respective tangents in some line parallel with the x-axis `("compare " 2/3 " and " -2/3)`.

This suggested that the point of intersection of the tangents should lie on the horizontal line through the centre of the circle (that is, through `y = 1/2`).

The current answer satisfies this requirement. It also suggests an alternative method for calculating the angle between the tangents; it will be twice the angle between one of the tangents and the horizontal. The angle between one of the tangents and the horizontal is the arctan of its slope. Choosing the tangent with positive slope for convenience, this would be `arctan(2/3) = 0.588` radians (three decimal places). The angle between the tangents is therefore `2 (0.588) = 1.176` (as previously noted but providing a check on the other method).