Circle Tangents and Their Equations

Following the Tangent When Leaving a Circular Path

Imagine riding around a circular track. If you leave the track without turning the handlebars, your direction at that instant follows the track's tangent line. This unique line passes through that point.

Mathematically, a tangent line meets a circle at exactly one point, called the point of tangency. The tangent is always perpendicular to the radius drawn to that point.

The perpendicular relationship is used in all three cases: constructing a tangent at a known point, finding tangents with a given slope, and drawing both tangents through a point outside the circle.

Tangent Line Through a Point on the Circle

Start by checking that the proposed point of tangency really lies on the circle. Its radius vector then becomes a normal vector for the tangent.

Tangent Line Through a Point on the Circle

The radius to the known contact point is normal to the tangent.

For the circle $(x-a)^2+(y-b)^2=r^2$ and point of tangency $T(x_1,y_1)$ , proceed as follows.

Step 1: Verify that $T$ lies on the circle.

(x_1-a)^2+(y_1-b)^2=r^2

Step 2: The radius vector is normal to the tangent.

\vec{OT}=(x_1-a,\,y_1-b)

Step 3: Use that normal vector in the line equation through $T$ .

\begin{aligned} &(x_1-a)(x-x_1) +(y_1-b)(y-y_1) \\ &=0 \end{aligned}

Using the point-on-circle condition from Step 1, this becomes the point-tangent form:

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

This derivation uses a normal vector, so the same steps work for vertical and horizontal tangents.

Tangent Line with Given Slope

If the point of tangency is unknown but the slope is given, the intercept must place the line exactly one radius away from the center. This often appears when the tangent must be parallel to another line.

Tangent Line with Given Slope

A given slope determines two parallel tangents, one on each side of the center.

For circle $x^2 + y^2 = r^2$ and known slope $m$ , we substitute the line equation $y = mx + c$ into the circle equation:

x^2 + (mx + c)^2 = r^2

Since the line is tangent to the circle, the discriminant of this quadratic equation must be zero:

\begin{aligned} \Delta &= (2mc)^2 - 4(1 + m^2)(c^2 - r^2) \\ &= 0 \end{aligned}

Solving for the intercept gives:

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

For every finite slope $m$ , the centered circle has two parallel tangent lines:

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

Tangents Through an Outside Point

A point outside a circle determines exactly two tangent lines. Their two contact points lie on one useful auxiliary line, the polar of the outside point.

Tangents Through a Point Outside the Circle

One outside point determines two contact points and two tangent lines.

Let $P(x_0,y_0)$ lie outside the circle $(x-a)^2+(y-b)^2=r^2$ . Its polar line is:

\begin{aligned} &(x-a)(x_0-a) + (y-b)(y_0-b) \\ &= r^2 \end{aligned}

For an outside point, this polar is the chord of contact, the line through both points of tangency. The construction is therefore:

Write the polar line.
Intersect it with the circle to find both points of tangency.
Use each contact point to construct one tangent through $P$ .

The polar is not itself one of the two tangents. It is the line joining their contact points.

Finding Two Tangents with the Polar Line

Find both tangents from $B(2,6)$ to the circle $L \equiv (x+1)^2+(y-2)^2=9$ .

Step 1: Determine the polar line of $B(2,6)$ with respect to the circle.

\begin{aligned} &(x-(-1))(2-(-1)) \\ &\quad{}+ (y-2)(6-2) = 9 \end{aligned}

Step 2: Find the intersection points of the polar line $3x + 4y = 14$ with the circle. From the line equation, we get $x = \frac{14-4y}{3}$ . Substitute into the circle equation:

(\frac{14-4y}{3} + 1)^2 + (y-2)^2 = 9

Step 3: Solve the quadratic for the two possible $y$ -coordinates.

(17-4y)^2+9(y-2)^2=81

Step 4: Substitute each value into the polar line to obtain the points of tangency.

x_1=\frac{14-4\left(\frac{122}{25}\right)}{3}=-\frac{46}{25}

Step 5: Use each point in the point-tangent form of the circle.

\begin{aligned} &-\frac{21}{25}(x+1)+\frac{72}{25}(y-2)=9 \\ &\Longrightarrow \\ &-7x+24y=130 \end{aligned}

Both lines pass through $B(2,6)$ and touch the circle once, so these are the required tangents.

Practice

Each problem asks whether a given line touches the circle and, if it does, where. Use the tangency condition to check the line before you locate the touch point.

Determine the equation of the tangent line to circle $x^2 + y^2 = 25$ at point $(3, 4)$ .
Find the equation of tangent lines to circle $x^2 + y^2 = 16$ that are parallel to line $y = 2x + 5$ .
Determine the equation of tangent lines to circle $(x-1)^2 + (y+2)^2 = 8$ passing through point $(4, 2)$ .
A circle has the equation $x^2 + y^2 - 4x + 6y - 3 = 0$ . Determine the equation of tangent lines that are perpendicular to line $x + 2y = 7$ .

Worked Solutions

Solution:

For circle $x^2 + y^2 = 25$ with center $O(0,0)$ and radius $r = 5$ , and tangent point $T(3,4)$ .

Using the tangent line formula: $(x-a)(x_1-a) + (y-b)(y_1-b) = r^2$
$\begin{aligned} &(x-0)(3-0) + (y-0)(4-0) \\ &= 25 \end{aligned}$
Therefore, the tangent line equation is $3x + 4y = 25$ .
Solution:

A line parallel to $y = 2x + 5$ has slope $m = 2$ .

For circle $x^2 + y^2 = 16$ with $r = 4$ , using the formula:

$y = mx \pm r\sqrt{1 + m^2}$
$y = 2x \pm 4\sqrt{1 + 4}$
Therefore, there are two tangent lines: $y = 2x + 4\sqrt{5}$ and $y = 2x - 4\sqrt{5}$ .
Solution:

Circle $(x-1)^2 + (y+2)^2 = 8$ has center $(1,-2)$ and $r^2 = 8$ .

Polar line equation for point $(4,2)$ :
$\begin{aligned} &(x-1)(4-1) \\ &\quad{}+ (y-(-2))(2-(-2)) = 8 \end{aligned}$
Translate the center to the origin with $u=x-1$ and $v=y+2$ . The circle and polar line become:
$u^2+v^2=8,\qquad 3u+4v=8$
Substituting back gives the two points of tangency:
$T_1\left(\frac{49-8\sqrt{34}}{25},\frac{-18+6\sqrt{34}}{25}\right)$
The slopes from $(4,2)$ to those points are $12-2\sqrt{34}$ and $12+2\sqrt{34}$ . Therefore, the required tangent lines are:
$y-2=(12-2\sqrt{34})(x-4)$
Solution:

Convert the circle equation to standard form by completing the square:
$x^2 + y^2 - 4x + 6y - 3 = 0$
Center $(2,-3)$ , radius $r = 4$ .

Line $x + 2y = 7$ can be written as $y = -\frac{1}{2}x + \frac{7}{2}$ , so its slope is $m_1 = -\frac{1}{2}$ .

A line perpendicular to it has slope $m = 2$ .

For the circle with center $(2,-3)$ , the tangent line equation with slope $2$ :
$y - (-3) = 2(x - 2) \pm 4\sqrt{1 + 4}$

The two tangent lines are $y = 2x - 7 + 4\sqrt{5}$ and $y = 2x - 7 - 4\sqrt{5}$ .

Following the Tangent When Leaving a Circular Path

Mathematically, a tangent line meets a circle at exactly one point, called the point of tangency. The tangent is always perpendicular to the radius drawn to that point.

The perpendicular relationship is used in all three cases: constructing a tangent at a known point, finding tangents with a given slope, and drawing both tangents through a point outside the circle.

Tangent Line Through a Point on the Circle

Start by checking that the proposed point of tangency really lies on the circle. Its radius vector then becomes a normal vector for the tangent.

Tangent Line Through a Point on the Circle

The radius to the known contact point is normal to the tangent.

For the circle $(x-a)^2+(y-b)^2=r^2$ and point of tangency $T(x_1,y_1)$ , proceed as follows.

Step 1: Verify that $T$ lies on the circle.

(x_1-a)^2+(y_1-b)^2=r^2

Step 2: The radius vector is normal to the tangent.

\vec{OT}=(x_1-a,\,y_1-b)

Step 3: Use that normal vector in the line equation through $T$ .

\begin{aligned} &(x_1-a)(x-x_1) +(y_1-b)(y-y_1) \\ &=0 \end{aligned}

Using the point-on-circle condition from Step 1, this becomes the point-tangent form:

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

This derivation uses a normal vector, so the same steps work for vertical and horizontal tangents.

Tangent Line with Given Slope

A given slope determines two parallel tangents, one on each side of the center.

For circle $x^2 + y^2 = r^2$ and known slope $m$ , we substitute the line equation $y = mx + c$ into the circle equation:

x^2 + (mx + c)^2 = r^2

Since the line is tangent to the circle, the discriminant of this quadratic equation must be zero:

\begin{aligned} \Delta &= (2mc)^2 - 4(1 + m^2)(c^2 - r^2) \\ &= 0 \end{aligned}

Solving for the intercept gives:

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

For every finite slope $m$ , the centered circle has two parallel tangent lines:

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

Tangents Through an Outside Point

A point outside a circle determines exactly two tangent lines. Their two contact points lie on one useful auxiliary line, the polar of the outside point.

Tangents Through a Point Outside the Circle

One outside point determines two contact points and two tangent lines.

Let $P(x_0,y_0)$ lie outside the circle $(x-a)^2+(y-b)^2=r^2$ . Its polar line is:

\begin{aligned} &(x-a)(x_0-a) + (y-b)(y_0-b) \\ &= r^2 \end{aligned}

For an outside point, this polar is the chord of contact, the line through both points of tangency. The construction is therefore:

Write the polar line.
Intersect it with the circle to find both points of tangency.
Use each contact point to construct one tangent through $P$ .

The polar is not itself one of the two tangents. It is the line joining their contact points.

Finding Two Tangents with the Polar Line

Find both tangents from $B(2,6)$ to the circle $L \equiv (x+1)^2+(y-2)^2=9$ .

Step 1: Determine the polar line of $B(2,6)$ with respect to the circle.

\begin{aligned} &(x-(-1))(2-(-1)) \\ &\quad{}+ (y-2)(6-2) = 9 \end{aligned}

Step 2: Find the intersection points of the polar line $3x + 4y = 14$ with the circle. From the line equation, we get $x = \frac{14-4y}{3}$ . Substitute into the circle equation:

(\frac{14-4y}{3} + 1)^2 + (y-2)^2 = 9

Step 3: Solve the quadratic for the two possible $y$ -coordinates.

(17-4y)^2+9(y-2)^2=81

Step 4: Substitute each value into the polar line to obtain the points of tangency.

x_1=\frac{14-4\left(\frac{122}{25}\right)}{3}=-\frac{46}{25}

Step 5: Use each point in the point-tangent form of the circle.

\begin{aligned} &-\frac{21}{25}(x+1)+\frac{72}{25}(y-2)=9 \\ &\Longrightarrow \\ &-7x+24y=130 \end{aligned}

Both lines pass through $B(2,6)$ and touch the circle once, so these are the required tangents.

Practice

Each problem asks whether a given line touches the circle and, if it does, where. Use the tangency condition to check the line before you locate the touch point.

Determine the equation of the tangent line to circle $x^2 + y^2 = 25$ at point $(3, 4)$ .
Find the equation of tangent lines to circle $x^2 + y^2 = 16$ that are parallel to line $y = 2x + 5$ .
Determine the equation of tangent lines to circle $(x-1)^2 + (y+2)^2 = 8$ passing through point $(4, 2)$ .
A circle has the equation $x^2 + y^2 - 4x + 6y - 3 = 0$ . Determine the equation of tangent lines that are perpendicular to line $x + 2y = 7$ .

Worked Solutions

Solution:

For circle $x^2 + y^2 = 25$ with center $O(0,0)$ and radius $r = 5$ , and tangent point $T(3,4)$ .

Using the tangent line formula: $(x-a)(x_1-a) + (y-b)(y_1-b) = r^2$
$\begin{aligned} &(x-0)(3-0) + (y-0)(4-0) \\ &= 25 \end{aligned}$
Therefore, the tangent line equation is $3x + 4y = 25$ .
Solution:

A line parallel to $y = 2x + 5$ has slope $m = 2$ .

For circle $x^2 + y^2 = 16$ with $r = 4$ , using the formula:

$y = mx \pm r\sqrt{1 + m^2}$
$y = 2x \pm 4\sqrt{1 + 4}$
Therefore, there are two tangent lines: $y = 2x + 4\sqrt{5}$ and $y = 2x - 4\sqrt{5}$ .
Solution:

Circle $(x-1)^2 + (y+2)^2 = 8$ has center $(1,-2)$ and $r^2 = 8$ .

Polar line equation for point $(4,2)$ :
$\begin{aligned} &(x-1)(4-1) \\ &\quad{}+ (y-(-2))(2-(-2)) = 8 \end{aligned}$
Translate the center to the origin with $u=x-1$ and $v=y+2$ . The circle and polar line become:
$u^2+v^2=8,\qquad 3u+4v=8$
Substituting back gives the two points of tangency:
$T_1\left(\frac{49-8\sqrt{34}}{25},\frac{-18+6\sqrt{34}}{25}\right)$
The slopes from $(4,2)$ to those points are $12-2\sqrt{34}$ and $12+2\sqrt{34}$ . Therefore, the required tangent lines are:
$y-2=(12-2\sqrt{34})(x-4)$
Solution:

Convert the circle equation to standard form by completing the square:
$x^2 + y^2 - 4x + 6y - 3 = 0$
Center $(2,-3)$ , radius $r = 4$ .

Line $x + 2y = 7$ can be written as $y = -\frac{1}{2}x + \frac{7}{2}$ , so its slope is $m_1 = -\frac{1}{2}$ .

A line perpendicular to it has slope $m = 2$ .

For the circle with center $(2,-3)$ , the tangent line equation with slope $2$ :
$y - (-3) = 2(x - 2) \pm 4\sqrt{1 + 4}$

The two tangent lines are $y = 2x - 7 + 4\sqrt{5}$ and $y = 2x - 7 - 4\sqrt{5}$ .