Tangent line to a circle

Geometry: Circles

Tangent line to a circle

When a line and a circle have exactly one point of intersection, the line is called a tangent to the circle.

Tangent to a circle

If a line $\blue l$ and circle $\green c$ have exactly one point of intersection $\orange T$ then $\blue l$ is called the tangent to $\green c$ at $\orange T$ .

We call the point $\orange T$ the tangent point.

In the figure, the point $\orange T$ can be moved across the circle $\green c$ . Circle $\green c$ can be moved by dragging its centre, and the radius can be adjusted using the slider.

Each tangent has the following property.

Tangent Theorem

A tangent to a circle is perpendicular to the line through its tangent point and the centre of the circle.

We will prove the theorem for a circle $\green c$ with centre $\rv{0,0}$ and radius $r$ , and a tangent $\blue l$ at point $\orange{T}=\orange{\rv{t,s}}$ on the circle. Moving the centre has no effect, the statement remains valid.

The circle equation of $\green c$ is equal to $x^2+y^2=r^2$ . By means of conversion, we can describe the circle by two graphs of a function, namely $f(x)=\sqrt{r^2-x^2}$ for the positive half of the circle, and $g(x)=-\sqrt{r^2-x^2}$ for the negative half of the circle.

Now, using differentiation, we can calculate the slope of the circle in the point $\orange{T}=\orange{\rv{t,s}}$ . The derivative of $f(x)$ is equal to $f'(x)=\frac{-2x}{2\sqrt{r^2-x^2}}=\frac{-x}{\sqrt{r^2-x^2}}$ . This means that the slope $\orange{T}=\orange{\rv{t,s}}$ equals:

$f'(t)=\frac{-t}{\sqrt{r^2-t^2}}$

Because the point $\orange{T}=\orange{\rv{t,s}}$ lies on the circle, we have $t^2+s^2=r^2$ This means that $r^2-t^2=s^2$ , and thus the following applies:

$f'(t)=\frac{-t}{\sqrt{s^2}}=-\frac{t}{s}$

Similarly, we can do this for $g(x)$ . Pay attention to the minus signs. Because $s$ is negative, the result is the same.

This means that the slope of line $\blue l$ must be equal to $-\frac{t}{s}$ . Otherwise, $\blue l$ is not a tangent.

Now we calculate the slope of the line $O\orange T$ , the radius of the circle through the point $\orange{T}=\orange{\rv{t,s}}$ .

The slope is equal to: $a_{O\orange T}=\frac{y_T-y_O}{x_T-x_O}=\frac{s-0}{t-0}=\frac{s}{t}$

We know that two lines are perpendicular to each other when the product of their slopes equals $-1$ .

In this case, for the slopes of line $\blue l$ and line $O\orange T$ , we have:

$a_{\blue l} \cdot a_{O\orange T}=-\frac{t}{s} \cdot \frac{s}{t}=-1$

This proves that the tangent $\blue l$ to circle $\green c$ at point $\orange{T}=\orange{\rv{t,s}}$ is perpendicular to the line through its tangent point and the circle's centre (the radius through the tangent point).

We can use the tangent theorem to determine an equation of a tangent to a circle through a given point.

Tangent to a circle

	Step-by-step	Example
	We determine an equation for a tangent line $\blue l$ to a circle $\green c$ at a point $\orange T$ .	$\orange T=\orange{\rv{5,7}}$ $\green c : \green{(x-2)^2+(y-3)^2=25}$
Step 1	Determine the centre $M$ of circle $\green c$ .	$M=\rv{2,3}$
Step 2	Determine the slope of the line through $M$ and $\orange T$ .	${a}_{M\orange T}=\tfrac{7-3}{5-2}=\tfrac{4}{3}$
Step 3	Use the tangent theorem to calculate the slope $a$ of tangent line $\blue l$ .	$a=-\tfrac{3}{4}$
step 4	The equation of line $\blue l$ is of the form $y=ax+b$ . Substitute the $a$ found in step 3.	$\blue l: y=-\tfrac{3}{4}x+b$
step 5	Determine $b$ by substituting the point $\orange T$ in the equation found in step 4 and solve the resulting equation.	$b=\tfrac{43}{4}$
step 6	Substitute $b$ into the equation of step 4. This gives an equation for line $\blue l$ .	$\blue l: y=-\tfrac{3}{4}x+\tfrac{43}{4}$

Picture

In the figure we see circle $\green c : \green{(x-2)^2+(y-3)^2=25}$ and point $\orange T=\orange{\rv{5,7}}$ with the tangent line $\blue l: y=-\tfrac{3}{4}x+\tfrac{43}{4}$ from the example.

Is line $l: {{35\cdot x}\over{2}}-5\cdot y=190$ a tangent to circle $c: \left(x+1\right)^2+\left(y-1\right)^2=85$ ?

Line $l$ is a tangent to circle $c$ when $l$ and circle $c$ have exactly one intersection point. We therefore determine the number of intersections of line $l$ and circle $c$ .

Step 1	We rewrite line $l$ to form $y=\ldots$ . That gives: $l: y={{7\cdot x}\over{2}}-38$
Step 2	We substitute the equation of line $l$ in the equation of the circle. That gives: $\left(x+1\right)^2+\left({{7\cdot x}\over{2}}-38-1\right)^2=85$ This can be simplified to: $\left({{7\cdot x}\over{2}}-39\right)^2+\left(x+1\right)^2=85$
Step 3	We reduce the equation from step 2 to $0$ and expand the brackets. This goes as follows: $\begin{array}{rcl}\left({{7\cdot x}\over{2}}-39\right)^2+\left(x+1\right)^2&=&85 \\&&\phantom{xxx}\blue{\text{original equation}} \\ {{53\cdot x^2}\over{4}}-271\cdot x+1522&=&85\\&&\phantom{xxx}\blue{\text{expanded brackets}} \\ {{53\cdot x^2}\over{4}}-271\cdot x+1437 &=& 0 \\&&\phantom{xxx}\blue{\text{reduced to }0} \ \end{array}$ We now read $a$ , $b$ and $c$ for the quadratic formula. That gives: $a={{53}\over{4}}$ , $b=-271$ and $c=1437$ . We can now calculate the discriminant. $\begin{array}{rcl}D&=&b^2-4ac \\&&\phantom{xxx}\blue{\text{formula discriminant}} \\ &=& (-271)^2-4\cdot {{53}\over{4}} \cdot 1437 \\&&\phantom{xxx}\blue{\text{substituted}} \\ &=& -2720 \end{array}$ Since the discriminant is equal to $-2720 \lt 0$ , there are there $0$ solutions .

Because there are $0$ intersection points, line $l$ is no tangent to $c$ .

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