Equation of a Tangent Line to a Circle

How the Tangent Meets the Radius

Picture a bicycle wheel resting on a level road. At that instant, the road touches the circular rim at exactly one point. The road itself models the tangent line, and the radius to the point of contact is perpendicular to it.

We will build the equation of that line in three situations: when the point of tangency is known, when the slope is known, and when the line must pass through a point outside the circle.

The geometry is the same in each situation. A tangent touches the circle once, while the radius to the point of tangency supplies the perpendicular direction or the required distance.

Tangent at a Known Point

When the point of tangency is known, first verify that the point lies on the circle. Then the radius to that point becomes a normal vector for the tangent.

Tangent at a Known Point

Constructing the tangent line equation when the point of tangency is known.

For the circle $x^2 + y^2 = r^2$ and a point of tangency $M(p, q)$ on that circle, the point-tangent form is:

px + qy = r^2

Why does this formula work? The radius $OM$ has direction vector $(p,q)$ . Because the tangent is perpendicular to $OM$ , that same vector is normal to the tangent. Substituting $M$ gives $p^2+q^2=r^2$ , so the line really does pass through the point of tangency.

For the circle $(x-h)^2 + (y-k)^2 = r^2$ with point of tangency $M(p,q)$ , translate the same idea to center $(h,k)$ :

\begin{aligned} &(p-h)(x-h) + (q-k)(y-k) \\ &= r^2 \end{aligned}

This form uses the radius directly as a normal vector. The formula also works for a vertical tangent, which has no finite slope.

Distance Method for a Given Slope

If the slope is known but the intercept is not, use the defining distance condition: the perpendicular distance from the center to a tangent equals the radius.

Tangents with a Given Slope

The center-to-line distance determines the two possible intercepts.

Suppose the circle is $(x-h)^2 + (y-k)^2 = r^2$ and the tangent has slope $m$ . Write the line as $y=mx+c$ .

The point-to-line distance formula gives the distance from center $(h,k)$ to line $mx-y+c=0$ :

d = \frac{|mh - k + c|}{\sqrt{m^2 + 1}}

For tangency, this distance must equal the radius:

\frac{|mh - k + c|}{\sqrt{m^2 + 1}} = r

Solving the absolute-value equation gives two intercepts:

c = k - mh \pm r\sqrt{m^2 + 1}

The two parallel tangents are:

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

Tangents Through an Outside Point

From a point outside a circle, exactly two tangent lines can be drawn. When both tangents have finite slopes, their slopes can be found from one quadratic equation.

Tangents Through an Outside Point

The repeated-intersection condition determines the two possible slopes.

Let the outside point be $P(x_0,y_0)$ and the circle be $x^2+y^2=r^2$ . A nonvertical line through $P$ has slope $m$ .

Its equation is $y-y_0=m(x-x_0)$ , or $y=mx-mx_0+y_0$ .

Substitute this expression into the circle. A tangent meets the circle at one repeated point, so the resulting quadratic in $x$ must have discriminant zero:

x^2 + (mx - mx_0 + y_0)^2 = r^2

After simplification, the discriminant condition becomes:

\begin{aligned} &(x_0^2 - r^2)m^2 - 2x_0y_0m + (y_0^2 - r^2) \\ &= 0 \end{aligned}

Its roots are the slopes of the two nonvertical tangents. If a vertical tangent is possible, handle $x=x_0$ separately because it cannot be written as $y=mx+c$ .

Finding Two Tangents from an Exterior Point

Determine the two tangent lines to $x^2+y^2=25$ that pass through $A(7,1)$ .

Step 1: Check that $A$ lies outside the circle.

7^2 + 1^2 = 49 + 1 = 50 > 25 \quad \checkmark

Step 2: Substitute $x_0=7$ , $y_0=1$ , and $r^2=25$ into the quadratic for $m$ :

\begin{aligned} &(49 - 25)m^2 - 2(7)(1)m + (1 - 25) \\ &= 0 \end{aligned}

Step 3: Solve the quadratic.

\begin{aligned} m &= \frac{7 \pm \sqrt{49 + 576}}{24} \\ &= \frac{7 \pm \sqrt{625}}{24} \\ &= \frac{7 \pm 25}{24} \end{aligned}

So $m_1 = \frac{32}{24} = \frac{4}{3}$ and $m_2 = \frac{-18}{24} = -\frac{3}{4}$ .

Step 4: Use each slope in the point-slope equation through $A$ .

y - 1 = \frac{4}{3}(x - 7) \Rightarrow 4x - 3y = 25

Practice

Each problem asks for the equation of a line that touches the circle at one point, so start from the tangency condition and then use the given point or gradient.

Determine the tangent line equation for circle $x^2 + y^2 = 16$ at point $(2\sqrt{2}, 2\sqrt{2})$ .
Find the tangent line equations for circle $x^2 + y^2 = 9$ with slope $-\frac{1}{2}$ .
Determine the tangent line equations for circle $(x-3)^2 + (y-4)^2 = 25$ that pass through point $(11, 10)$ .
A circle has equation $x^2 + y^2 - 6x - 8y = 0$ . Determine the tangent line equations parallel to line $3x + 4y = 7$ .

Worked Solutions

Solution:

For circle $x^2 + y^2 = 16$ at point $(2\sqrt{2}, 2\sqrt{2})$ , use the formula:

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

Step 1: Substitute tangent point coordinates
$x \cdot 2\sqrt{2} + y \cdot 2\sqrt{2} = 16$
Step 2: Simplify by dividing both sides by $2\sqrt{2}$

$\begin{aligned} x + y &= \frac{16}{2\sqrt{2}} \\ &= \frac{8}{\sqrt{2}} \\ &= \frac{8\sqrt{2}}{2} \\ &= 4\sqrt{2} \end{aligned}$

So the tangent line equation is $x + y = 4\sqrt{2}$ .
Solution:

For gradient $m = -\frac{1}{2}$ and $r = 3$ , use the formula:

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

Step 1: Calculate value of $\sqrt{1 + m^2}$

$\begin{aligned} \sqrt{1 + m^2} &= \sqrt{1 + \left(-\frac{1}{2}\right)^2} \\ &= \sqrt{1 + \frac{1}{4}} \\ &= \sqrt{\frac{5}{4}} \\ &= \frac{\sqrt{5}}{2} \end{aligned}$

Step 2: Substitute into formula

$y = -\frac{1}{2}x \pm 3 \cdot \frac{\sqrt{5}}{2} = -\frac{1}{2}x \pm \frac{3\sqrt{5}}{2}$

So the two tangent line equations are: $y = -\frac{1}{2}x + \frac{3\sqrt{5}}{2}$ and $y = -\frac{1}{2}x - \frac{3\sqrt{5}}{2}$ .
Solution:

Circle $(x-3)^2 + (y-4)^2 = 25$ has center $(3,4)$ and radius $r = 5$ .

Step 1: Check position of point $(11,10)$

$\begin{aligned} &(11-3)^2 + (10-4)^2 \\ &= 8^2 + 6^2 \\ &= 64 + 36 \\ &= 100 > 25 \end{aligned}$

Point is outside the circle.

Step 2: Translate the center to the origin. Relative to the center, point $(11,10)$ has coordinates $(8,6)$ .

Step 3: Form the quadratic for the slopes from $(8,6)$ to circle $x^2 + y^2 = 25$ .
$\begin{aligned} &(8^2 - 25)m^2 - 2(8)(6)m \\ &\quad{}+ (6^2 - 25) = 0 \end{aligned}$
Step 4: Solve it with the quadratic formula.
$\begin{aligned} m &= \frac{96 \pm \sqrt{96^2 - 4(39)(11)}}{2(39)} \\ &= \frac{96 \pm \sqrt{9216 - 1716}}{78} \end{aligned}$
Step 5: Simplify the slopes and use point-slope form through $(11,10)$ .
$\begin{aligned} &m_1 = \frac{48 + 25\sqrt{3}}{39} \\ &\text{and} \\ &m_2 = \frac{48 - 25\sqrt{3}}{39} \end{aligned}$
These are the two tangent lines through $(11,10)$ .
Solution:

Step 1: Complete the square to write the circle in standard form.
$x^2 + y^2 - 6x - 8y = 0$
Center $(3,4)$ , radius $r = 5$ .

Step 2: Read the slope of the given line. Since $3x + 4y = 7$ is $y = -\frac{3}{4}x + \frac{7}{4}$ , every parallel line has slope $m = -\frac{3}{4}$ .

Step 3: Substitute center $(3,4)$ , radius, and slope into the tangent formula.
$y - 4 = -\frac{3}{4}(x - 3) \pm 5\sqrt{1 + \frac{9}{16}}$
Step 4: Simplify both equations.
$y = -\frac{3}{4}x + \frac{50}{4} = -\frac{3}{4}x + \frac{25}{2}$
So the two tangent line equations are: $y = -\frac{3}{4}x + \frac{25}{2}$ and $y = -\frac{3}{4}x$ .

How the Tangent Meets the Radius

We will build the equation of that line in three situations: when the point of tangency is known, when the slope is known, and when the line must pass through a point outside the circle.

The geometry is the same in each situation. A tangent touches the circle once, while the radius to the point of tangency supplies the perpendicular direction or the required distance.

Tangent at a Known Point

When the point of tangency is known, first verify that the point lies on the circle. Then the radius to that point becomes a normal vector for the tangent.

Tangent at a Known Point

Constructing the tangent line equation when the point of tangency is known.

For the circle $x^2 + y^2 = r^2$ and a point of tangency $M(p, q)$ on that circle, the point-tangent form is:

px + qy = r^2

For the circle $(x-h)^2 + (y-k)^2 = r^2$ with point of tangency $M(p,q)$ , translate the same idea to center $(h,k)$ :

\begin{aligned} &(p-h)(x-h) + (q-k)(y-k) \\ &= r^2 \end{aligned}

This form uses the radius directly as a normal vector. The formula also works for a vertical tangent, which has no finite slope.

Distance Method for a Given Slope

If the slope is known but the intercept is not, use the defining distance condition: the perpendicular distance from the center to a tangent equals the radius.

Tangents with a Given Slope

The center-to-line distance determines the two possible intercepts.

Suppose the circle is $(x-h)^2 + (y-k)^2 = r^2$ and the tangent has slope $m$ . Write the line as $y=mx+c$ .

The point-to-line distance formula gives the distance from center $(h,k)$ to line $mx-y+c=0$ :

d = \frac{|mh - k + c|}{\sqrt{m^2 + 1}}

For tangency, this distance must equal the radius:

\frac{|mh - k + c|}{\sqrt{m^2 + 1}} = r

Solving the absolute-value equation gives two intercepts:

c = k - mh \pm r\sqrt{m^2 + 1}

The two parallel tangents are:

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

Tangents Through an Outside Point

From a point outside a circle, exactly two tangent lines can be drawn. When both tangents have finite slopes, their slopes can be found from one quadratic equation.

Tangents Through an Outside Point

The repeated-intersection condition determines the two possible slopes.

Let the outside point be $P(x_0,y_0)$ and the circle be $x^2+y^2=r^2$ . A nonvertical line through $P$ has slope $m$ .

Its equation is $y-y_0=m(x-x_0)$ , or $y=mx-mx_0+y_0$ .

Substitute this expression into the circle. A tangent meets the circle at one repeated point, so the resulting quadratic in $x$ must have discriminant zero:

x^2 + (mx - mx_0 + y_0)^2 = r^2

After simplification, the discriminant condition becomes:

\begin{aligned} &(x_0^2 - r^2)m^2 - 2x_0y_0m + (y_0^2 - r^2) \\ &= 0 \end{aligned}

Its roots are the slopes of the two nonvertical tangents. If a vertical tangent is possible, handle $x=x_0$ separately because it cannot be written as $y=mx+c$ .

Finding Two Tangents from an Exterior Point

Determine the two tangent lines to $x^2+y^2=25$ that pass through $A(7,1)$ .

Step 1: Check that $A$ lies outside the circle.

7^2 + 1^2 = 49 + 1 = 50 > 25 \quad \checkmark

Step 2: Substitute $x_0=7$ , $y_0=1$ , and $r^2=25$ into the quadratic for $m$ :

\begin{aligned} &(49 - 25)m^2 - 2(7)(1)m + (1 - 25) \\ &= 0 \end{aligned}

Step 3: Solve the quadratic.

\begin{aligned} m &= \frac{7 \pm \sqrt{49 + 576}}{24} \\ &= \frac{7 \pm \sqrt{625}}{24} \\ &= \frac{7 \pm 25}{24} \end{aligned}

So $m_1 = \frac{32}{24} = \frac{4}{3}$ and $m_2 = \frac{-18}{24} = -\frac{3}{4}$ .

Step 4: Use each slope in the point-slope equation through $A$ .

y - 1 = \frac{4}{3}(x - 7) \Rightarrow 4x - 3y = 25

Practice

Each problem asks for the equation of a line that touches the circle at one point, so start from the tangency condition and then use the given point or gradient.

Determine the tangent line equation for circle $x^2 + y^2 = 16$ at point $(2\sqrt{2}, 2\sqrt{2})$ .
Find the tangent line equations for circle $x^2 + y^2 = 9$ with slope $-\frac{1}{2}$ .
Determine the tangent line equations for circle $(x-3)^2 + (y-4)^2 = 25$ that pass through point $(11, 10)$ .
A circle has equation $x^2 + y^2 - 6x - 8y = 0$ . Determine the tangent line equations parallel to line $3x + 4y = 7$ .

Worked Solutions

Solution:

For circle $x^2 + y^2 = 16$ at point $(2\sqrt{2}, 2\sqrt{2})$ , use the formula:

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

Step 1: Substitute tangent point coordinates
$x \cdot 2\sqrt{2} + y \cdot 2\sqrt{2} = 16$
Step 2: Simplify by dividing both sides by $2\sqrt{2}$

$\begin{aligned} x + y &= \frac{16}{2\sqrt{2}} \\ &= \frac{8}{\sqrt{2}} \\ &= \frac{8\sqrt{2}}{2} \\ &= 4\sqrt{2} \end{aligned}$

So the tangent line equation is $x + y = 4\sqrt{2}$ .
Solution:

For gradient $m = -\frac{1}{2}$ and $r = 3$ , use the formula:

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

Step 1: Calculate value of $\sqrt{1 + m^2}$

$\begin{aligned} \sqrt{1 + m^2} &= \sqrt{1 + \left(-\frac{1}{2}\right)^2} \\ &= \sqrt{1 + \frac{1}{4}} \\ &= \sqrt{\frac{5}{4}} \\ &= \frac{\sqrt{5}}{2} \end{aligned}$

Step 2: Substitute into formula

$y = -\frac{1}{2}x \pm 3 \cdot \frac{\sqrt{5}}{2} = -\frac{1}{2}x \pm \frac{3\sqrt{5}}{2}$

So the two tangent line equations are: $y = -\frac{1}{2}x + \frac{3\sqrt{5}}{2}$ and $y = -\frac{1}{2}x - \frac{3\sqrt{5}}{2}$ .
Solution:

Circle $(x-3)^2 + (y-4)^2 = 25$ has center $(3,4)$ and radius $r = 5$ .

Step 1: Check position of point $(11,10)$

$\begin{aligned} &(11-3)^2 + (10-4)^2 \\ &= 8^2 + 6^2 \\ &= 64 + 36 \\ &= 100 > 25 \end{aligned}$

Point is outside the circle.

Step 2: Translate the center to the origin. Relative to the center, point $(11,10)$ has coordinates $(8,6)$ .

Step 3: Form the quadratic for the slopes from $(8,6)$ to circle $x^2 + y^2 = 25$ .
$\begin{aligned} &(8^2 - 25)m^2 - 2(8)(6)m \\ &\quad{}+ (6^2 - 25) = 0 \end{aligned}$
Step 4: Solve it with the quadratic formula.
$\begin{aligned} m &= \frac{96 \pm \sqrt{96^2 - 4(39)(11)}}{2(39)} \\ &= \frac{96 \pm \sqrt{9216 - 1716}}{78} \end{aligned}$
Step 5: Simplify the slopes and use point-slope form through $(11,10)$ .
$\begin{aligned} &m_1 = \frac{48 + 25\sqrt{3}}{39} \\ &\text{and} \\ &m_2 = \frac{48 - 25\sqrt{3}}{39} \end{aligned}$
These are the two tangent lines through $(11,10)$ .
Solution:

Step 1: Complete the square to write the circle in standard form.
$x^2 + y^2 - 6x - 8y = 0$
Center $(3,4)$ , radius $r = 5$ .

Step 2: Read the slope of the given line. Since $3x + 4y = 7$ is $y = -\frac{3}{4}x + \frac{7}{4}$ , every parallel line has slope $m = -\frac{3}{4}$ .

Step 3: Substitute center $(3,4)$ , radius, and slope into the tangent formula.
$y - 4 = -\frac{3}{4}(x - 3) \pm 5\sqrt{1 + \frac{9}{16}}$
Step 4: Simplify both equations.
$y = -\frac{3}{4}x + \frac{50}{4} = -\frac{3}{4}x + \frac{25}{2}$
So the two tangent line equations are: $y = -\frac{3}{4}x + \frac{25}{2}$ and $y = -\frac{3}{4}x$ .