The Circle — Tangents & Points

Section 1 of 2

The Slope of a Tangent

A tangent touches the circle at one point and is perpendicular to the radius there. Find the radius’s slope, flip it, and you have the tangent’s slope.

Three steps

1. Slope of the radius from $(0,0)$ to $(a,b)$: $m_r = \dfrac{b}{a}$. 2. Slope of the tangent (perpendicular): $m_t = -\dfrac{a}{b}$ (flip the fraction, change the sign). 3. Line: $y - y_1 = m(x - x_1)$.

Worked example — whole-number slope

Find the tangent to $x^2 + y^2 = 5$ at $(2, 1)$.

$m_r = \tfrac{1}{2}$, so $m_t = -\tfrac{2}{1} = -2$

$y - 1 = -2(x - 2) = -2x + 4$

$2x + y = 5$

Worked example — fraction slope — clear it

Find the tangent to $x^2 + y^2 = 13$ at $(3, 2)$.

$m_r = \tfrac{2}{3}$, so $m_t = -\tfrac{3}{2}$

$y - 2 = -\tfrac{3}{2}(x - 3)$, multiply across: $2(y-2) = -3(x-3)$

$2y - 4 = -3x + 9$

$3x + 2y = 13$

You try

The point $(3, -6)$ lies on $x^2 + y^2 = k$. Find $k$ and the tangent at that point.

First $r^2 = 3^2 + (-6)^2$. Then $m_r = \tfrac{-6}{3}$, flip and change sign for $m_t$.

$r^2 = 9 + 36 = 45$, so $x^2 + y^2 = 45$

$m_r = \tfrac{-6}{3} = -2$, so $m_t = \tfrac{1}{2}$

$y + 6 = \tfrac{1}{2}(x - 3) \Rightarrow 2(y+6) = x - 3$

$k = 45$; tangent $x - 2y = 15$

Section 2 of 2

Inside, On or Outside & Missing Coordinate

Position of a point

Substitute $(x_1, y_1)$ into $x^2 + y^2$ and compare with $r^2$: $< r^2 \Rightarrow$ inside; $= r^2 \Rightarrow$ on; $> r^2 \Rightarrow$ outside.

Worked example — inside or outside

Is $(3, -7)$ inside, on or outside $x^2 + y^2 = 59$? And $(-5, -9)$ on $x^2 + y^2 = 105$?

$(3)^2 + (-7)^2 = 9 + 49 = 58 < 59 \Rightarrow$ inside

$(-5)^2 + (-9)^2 = 25 + 81 = 106 > 105 \Rightarrow$ outside

$(3,-7)$ inside; $(-5,-9)$ outside

Find a missing coordinate

If the point is on the circle, slot it into the equation and solve. When you square-root, always write $\pm$ — there are usually two points.

Worked example — missing coordinate

The point $(2, k)$ lies on $x^2 + y^2 = 20$. Find the two values of $k$.

$(2)^2 + k^2 = 20$

$4 + k^2 = 20 \Rightarrow k^2 = 16$

$k = \pm\sqrt{16} = \pm 4$

$k = 4$ or $k = -4$

You try

The point $(2, -6)$ lies on $x^2 + y^2 = k$. Find $k$ and the tangent at that point.

$r^2 = 2^2 + (-6)^2$. Then $m_r = \tfrac{-6}{2}$, flip and change sign.

$r^2 = 4 + 36 = 40$

$m_r = -3$, so $m_t = \tfrac{1}{3}$; $y + 6 = \tfrac{1}{3}(x - 2) \Rightarrow 3(y+6) = x - 2$

$k = 40$; tangent $x - 3y = 20$

That’s Part 2.

Tangents (perpendicular to the radius), telling if a point is inside/on/outside, and finding a missing coordinate. Part 3: circles centred at $(h, k)$.