Show that the line with equation $y=3x-5$ is a tangent to the circle with equation $x^2 + y^2 + 2x - 4y - 5 = 0$ and find the coordinates of the point of contact. - Scottish Highers Maths - Question 8 - 2016

Next, we substitute the line's equation $y = 3x - 5$ into the circle's equation. Thus:

$(x+1)^2 + ((3x-5)-2)^2 = 10$

This simplifies to:

$(x+1)^2 + (3x-7)^2 = 10$

Expanding the terms:

$(x^2 + 2x + 1) + (9x^2 - 42x + 49) = 10$

Combining like terms results in:

$10x^2 - 40x + 40 = 0$

We can simplify further:

$x^2 - 4x + 4 = 0$

Next, we apply the discriminant to check for tangency:

$b^2 - 4ac = (-4)^2 - 4(1)(4) = 16 - 16 = 0$

Since the discriminant equals zero, it indicates one solution, which confirms the line is tangent to the circle.

To find the point of contact, we solve the quadratic equation found earlier:

$x - 2 = 0$

Thus:

$x = 2$

Substituting $x = 2$ back into the line's equation:

$y = 3(2) - 5 = 1$

Therefore, the coordinates of the point of contact are:

$(2, 1)$.