Tangents to Circles

One Contact Point and a Perpendicular Radius

A circle tangent line is a line that intersects the circle at exactly one point. The intersection point between the tangent line and the circle is called the point of tangency.

Circle Tangent Line

A tangent line intersects the circle at exactly one point.

A tangent line is always perpendicular to the radius at the point of tangency.

Equation of Circle Tangent Line

A tangent touches the circle at exactly one point, and the radius to that point is perpendicular to the tangent. That perpendicular condition supplies the slope of the tangent line, because the two slopes multiply to $-1$ .

Tangent Line Through a Point on the Circle

If point $(x_1, y_1)$ lies on the circle $x^2 + y^2 = r^2$ , then the equation of the tangent line at that point is:

x_1 \cdot x + y_1 \cdot y = r^2

The circle in this formula uses the standard form for a circle centered at the origin.

For a circle with center $(a, b)$ :

\begin{aligned} &(x_1 - a)(x - a) + (y_1 - b)(y - b) \\ &= r^2 \end{aligned}

Tangent Line with a Given Gradient

The equation of the tangent line to circle $x^2 + y^2 = r^2$ with gradient $m$ is:

y = mx \pm r\sqrt{1 + m^2}

For a circle with center $(a, b)$ :

y - b = m(x - a) \pm r\sqrt{1 + m^2}

Tangent Lines from an External Point

From a point outside the circle, two tangent lines can be drawn to the circle.

Two Tangent Lines from External Point

From point

P

outside the circle, two tangent lines can be drawn.

Length of Tangent Line

If $P(x_1, y_1)$ is a point outside the circle with center $O(a, b)$ and radius $r$ , then the length of the tangent segment from $P$ to the point of tangency is:

PT = \sqrt{(x_1 - a)^2 + (y_1 - b)^2 - r^2}

Common Tangent Lines of Two Circles

Two circles can share tangent lines that cross between them or pass outside. Whether a common tangent exists at all depends on how far apart the two circles are.

External Common Tangent Lines

External common tangent lines touch both circles without crossing the segment between their centers.

External Common Tangent Lines

Lines that touch both circles from the same side.

Length of external common tangent line:

l = \sqrt{d^2 - (r_1 - r_2)^2}

In this formula, $d$ is the distance between the two circle centers.

This positive-length external tangent segment exists only when $d>|r_1-r_2|$ . Equality describes internal tangency, where the segment between the two contact points collapses to length zero.

Internal Common Tangent Lines

Internal common tangent lines are lines that touch both circles and intersect the line connecting the two circle centers.

Internal Common Tangent Lines

Lines that touch both circles from opposite sides.

Length of internal common tangent line:

l = \sqrt{d^2 - (r_1 + r_2)^2}

This positive-length internal tangent segment exists only when $d>r_1+r_2$ . At equality, the circles touch externally and the segment between the contact points has length zero.

Determining Tangent Line Equations

The examples below build tangent equations from different starting information. Work through the substitution in each case. Check at the end of each example that the point of contact really lies on both the line and the circle.

Determining Tangent Line Through a Point on the Circle

Find the equation of the tangent line to circle $x^2 + y^2 = 25$ at point $(3, 4)$ .

Solution:

Since point $(3, 4)$ lies on the circle (can be verified: $3^2 + 4^2 = 9 + 16 = 25$ ), the equation of the tangent line is:

x_1 \cdot x + y_1 \cdot y = r^2

Determining Tangent Line with a Given Gradient

Find the equation of the tangent line to circle $x^2 + y^2 = 16$ that is parallel to line $3x - 4y + 5 = 0$ .

Solution:

The gradient of line $3x - 4y + 5 = 0$ is $m = \frac{3}{4}$ .

Equation of tangent line with gradient $m = \frac{3}{4}$ :

y = mx \pm r\sqrt{1 + m^2}

The equations of the tangent lines are:

$y = \frac{3}{4}x + 5$ or $3x - 4y + 20 = 0$
$y = \frac{3}{4}x - 5$ or $3x - 4y - 20 = 0$

Calculating the Length of Tangent Line from External Point

Find the length of the tangent line from point $P(7, 1)$ to circle $x^2 + y^2 = 25$ .

Solution:

Circle center $O(0, 0)$ and radius $r = 5$ .

PT = \sqrt{(x_1 - a)^2 + (y_1 - b)^2 - r^2}

Practice Problems

Each problem gives a circle and an extra condition such as a point or a parallel line. Use the perpendicular radius to build the tangent equation.

Find the equation of the tangent line to circle $x^2 + y^2 = 36$ at point $(-3, 3\sqrt{3})$ !
Find the equation of the tangent line to circle $(x - 2)^2 + (y + 3)^2 = 25$ that is perpendicular to line $2x + y - 5 = 0$ !
From point $A(10, 0)$ tangent lines are drawn to circle $x^2 + y^2 = 36$ . Find:
- Length of tangent line
- Coordinates of tangent points
Two circles are centered at $O_1(-4, 0)$ with radius $2$ and $O_2(4, 0)$ with radius $3$ . Find the length of the external common tangent line!
Find the equation of the tangent line to circle $x^2 + y^2 - 6x + 4y - 12 = 0$ that passes through point $(8, -2)$ !

Worked Solutions

Tangent line equation at a point on the circle

Verify point on circle:

$(-3)^2 + (3\sqrt{3})^2 = 9 + 27 = 36 \space \checkmark$

Tangent line equation:
$x_1 \cdot x + y_1 \cdot y = r^2$
Tangent line perpendicular to a given line

Gradient of line $2x + y - 5 = 0$ is $m_1 = -2$ .

Since perpendicular, then $m_2 = \frac{1}{2}$ .

Tangent line equation:
$y - b = m(x - a) \pm r\sqrt{1 + m^2}$
Therefore: $x - 2y - 8 \pm 5\sqrt{5} = 0$
Tangent line from external point
- Length of tangent line:
  $\begin{aligned} AT &= \sqrt{10^2 + 0^2 - 36} \\ &= \sqrt{100 - 36} \\ &= \sqrt{64} \\ &= 8 \end{aligned}$
- To find the tangent points, let $T(x,y)$ be one of them. Since radius $OT$ is perpendicular to tangent $AT$ , their direction vectors have zero dot product:
  $(A-T)\cdot T=0$
  Therefore, the tangent points are $T_1\left(\frac{18}{5},\frac{24}{5}\right)$ and $T_2\left(\frac{18}{5},-\frac{24}{5}\right)$ .
External common tangent line
$d = \sqrt{(4-(-4))^2 + (0-0)^2} = 8$
Tangent line through a point on the circle

Circle: $(x - 3)^2 + (y + 2)^2 = 25$

Center $(3, -2)$ , radius $r = 5$

Verify that point $(8, -2)$ lies on the circle:

$\begin{aligned} (8-3)^2 + (-2+2)^2 &= 25 + 0 \\ &= 25 \end{aligned}$

The left side equals $r^2=25$ , so the point lies on the circle. Therefore the tangent line equation is:
$\begin{aligned} &(8-3)(x-3) \\ &\quad{}+ (-2+2)(y+2) = 25 \end{aligned}$

One Contact Point and a Perpendicular Radius

A circle tangent line is a line that intersects the circle at exactly one point. The intersection point between the tangent line and the circle is called the point of tangency.

Circle Tangent Line

A tangent line intersects the circle at exactly one point.

A tangent line is always perpendicular to the radius at the point of tangency.

Equation of Circle Tangent Line

Tangent Line Through a Point on the Circle

If point $(x_1, y_1)$ lies on the circle $x^2 + y^2 = r^2$ , then the equation of the tangent line at that point is:

x_1 \cdot x + y_1 \cdot y = r^2

The circle in this formula uses the standard form for a circle centered at the origin.

For a circle with center $(a, b)$ :

\begin{aligned} &(x_1 - a)(x - a) + (y_1 - b)(y - b) \\ &= r^2 \end{aligned}

Tangent Line with a Given Gradient

The equation of the tangent line to circle $x^2 + y^2 = r^2$ with gradient $m$ is:

y = mx \pm r\sqrt{1 + m^2}

For a circle with center $(a, b)$ :

y - b = m(x - a) \pm r\sqrt{1 + m^2}

Tangent Lines from an External Point

From a point outside the circle, two tangent lines can be drawn to the circle.

Two Tangent Lines from External Point

From point

P

outside the circle, two tangent lines can be drawn.

Length of Tangent Line

If $P(x_1, y_1)$ is a point outside the circle with center $O(a, b)$ and radius $r$ , then the length of the tangent segment from $P$ to the point of tangency is:

PT = \sqrt{(x_1 - a)^2 + (y_1 - b)^2 - r^2}

Common Tangent Lines of Two Circles

Two circles can share tangent lines that cross between them or pass outside. Whether a common tangent exists at all depends on how far apart the two circles are.

External Common Tangent Lines

External common tangent lines touch both circles without crossing the segment between their centers.

External Common Tangent Lines

Lines that touch both circles from the same side.

Length of external common tangent line:

l = \sqrt{d^2 - (r_1 - r_2)^2}

In this formula, $d$ is the distance between the two circle centers.

This positive-length external tangent segment exists only when $d>|r_1-r_2|$ . Equality describes internal tangency, where the segment between the two contact points collapses to length zero.

Internal Common Tangent Lines

Internal common tangent lines are lines that touch both circles and intersect the line connecting the two circle centers.

Internal Common Tangent Lines

Lines that touch both circles from opposite sides.

Length of internal common tangent line:

l = \sqrt{d^2 - (r_1 + r_2)^2}

This positive-length internal tangent segment exists only when $d>r_1+r_2$ . At equality, the circles touch externally and the segment between the contact points has length zero.

Determining Tangent Line Equations

Determining Tangent Line Through a Point on the Circle

Find the equation of the tangent line to circle $x^2 + y^2 = 25$ at point $(3, 4)$ .

Solution:

Since point $(3, 4)$ lies on the circle (can be verified: $3^2 + 4^2 = 9 + 16 = 25$ ), the equation of the tangent line is:

x_1 \cdot x + y_1 \cdot y = r^2

Determining Tangent Line with a Given Gradient

Find the equation of the tangent line to circle $x^2 + y^2 = 16$ that is parallel to line $3x - 4y + 5 = 0$ .

Solution:

The gradient of line $3x - 4y + 5 = 0$ is $m = \frac{3}{4}$ .

Equation of tangent line with gradient $m = \frac{3}{4}$ :

y = mx \pm r\sqrt{1 + m^2}

The equations of the tangent lines are:

$y = \frac{3}{4}x + 5$ or $3x - 4y + 20 = 0$
$y = \frac{3}{4}x - 5$ or $3x - 4y - 20 = 0$

Calculating the Length of Tangent Line from External Point

Find the length of the tangent line from point $P(7, 1)$ to circle $x^2 + y^2 = 25$ .

Solution:

Circle center $O(0, 0)$ and radius $r = 5$ .

PT = \sqrt{(x_1 - a)^2 + (y_1 - b)^2 - r^2}

Practice Problems

Each problem gives a circle and an extra condition such as a point or a parallel line. Use the perpendicular radius to build the tangent equation.

Find the equation of the tangent line to circle $x^2 + y^2 = 36$ at point $(-3, 3\sqrt{3})$ !
Find the equation of the tangent line to circle $(x - 2)^2 + (y + 3)^2 = 25$ that is perpendicular to line $2x + y - 5 = 0$ !
From point $A(10, 0)$ tangent lines are drawn to circle $x^2 + y^2 = 36$ . Find:
- Length of tangent line
- Coordinates of tangent points
Two circles are centered at $O_1(-4, 0)$ with radius $2$ and $O_2(4, 0)$ with radius $3$ . Find the length of the external common tangent line!
Find the equation of the tangent line to circle $x^2 + y^2 - 6x + 4y - 12 = 0$ that passes through point $(8, -2)$ !

Worked Solutions

Tangent line equation at a point on the circle

Verify point on circle:

$(-3)^2 + (3\sqrt{3})^2 = 9 + 27 = 36 \space \checkmark$

Tangent line equation:
$x_1 \cdot x + y_1 \cdot y = r^2$
Tangent line perpendicular to a given line

Gradient of line $2x + y - 5 = 0$ is $m_1 = -2$ .

Since perpendicular, then $m_2 = \frac{1}{2}$ .

Tangent line equation:
$y - b = m(x - a) \pm r\sqrt{1 + m^2}$
Therefore: $x - 2y - 8 \pm 5\sqrt{5} = 0$
Tangent line from external point
- Length of tangent line:
  $\begin{aligned} AT &= \sqrt{10^2 + 0^2 - 36} \\ &= \sqrt{100 - 36} \\ &= \sqrt{64} \\ &= 8 \end{aligned}$
- To find the tangent points, let $T(x,y)$ be one of them. Since radius $OT$ is perpendicular to tangent $AT$ , their direction vectors have zero dot product:
  $(A-T)\cdot T=0$
  Therefore, the tangent points are $T_1\left(\frac{18}{5},\frac{24}{5}\right)$ and $T_2\left(\frac{18}{5},-\frac{24}{5}\right)$ .
External common tangent line
$d = \sqrt{(4-(-4))^2 + (0-0)^2} = 8$
Tangent line through a point on the circle

Circle: $(x - 3)^2 + (y + 2)^2 = 25$

Center $(3, -2)$ , radius $r = 5$

Verify that point $(8, -2)$ lies on the circle:

$\begin{aligned} (8-3)^2 + (-2+2)^2 &= 25 + 0 \\ &= 25 \end{aligned}$

The left side equals $r^2=25$ , so the point lies on the circle. Therefore the tangent line equation is:
$\begin{aligned} &(8-3)(x-3) \\ &\quad{}+ (-2+2)(y+2) = 25 \end{aligned}$