Construction of Tangents to a Circle

Construction of a Tangent to a Circle at a Point on the Circle (using the property that radius is perpendicular to tangent)

A tangent line to a circle in planar geometry is a straight line that touches the circle at exactly one point, without crossing into the circle's interior. The point where the tangent touches the circle is called the point of tangency (or point of contact).

A fundamental theorem in geometry states that the radius drawn to the point of tangency is always perpendicular to the tangent line at that point.

This property forms the basis for constructing a tangent at a given point on the circle.

Construction Steps

Given: A circle with center $O$ and a specific point $P$ located on the circumference of the circle.

Tools: Compass, Straightedge.

Goal: Construct the tangent line to the circle that passes through the given point $P$.

Steps:

1. Draw the Radius: Use the straightedge to draw the line segment connecting the center $O$ to the point $P$ on the circle. This segment is the radius $OP$.
2. Construct Perpendicular at P: The tangent line must be perpendicular to the radius $OP$ at point $P$. Construct a line that is perpendicular to the line containing $OP$ at the point $P$. Use the standard compass-and-straightedge method for constructing a perpendicular to a line from a point on the line (I1).
   • Extend the radius $OP$ slightly beyond $P$ if needed to facilitate drawing arcs.
   • Place the pointed end of the compass on point $P$. Using any convenient radius, draw arcs that intersect the line containing $OP$ on both sides of $P$. Label the intersection points $A$ and $B$. Note that $PA = PB$.
   • With $A$ as the center and a compass radius greater than $AP$, draw an arc above (or below) the line.
   • With $B$ as the center and the same compass width as in the previous step, draw another arc intersecting the arc drawn from $A$. Label the intersection point $Q$.
   • Use the straightedge to draw the line passing through points $P$ and $Q$. Extend this line as needed.

The line $PQ$ constructed in Step 2 is the required tangent line to the circle at point $P$.

Justification:

The construction directly implements the property that the tangent is perpendicular to the radius at the point of contact. We drew the radius $OP$ to the given point $P$ on the circle. We then constructed a line through $P$ that is perpendicular to $OP$. By the theorem "A line drawn through the endpoint of a radius and perpendicular to it is tangent to the circle", the constructed line must be the tangent at point $P$.

Example

Answer:

Construction of Tangents to a Circle from a Point outside the Circle

If you take a point outside a circle, you can draw exactly two distinct tangent lines from that point to the circle. These tangents will touch the circle at two different points of tangency. This construction finds those two points of tangency, and thus the tangent lines.

The construction is based on the property that the angle subtended by a diameter in a semicircle is a right angle, which relates to the tangent-radius property.

Construction Steps

Given: A circle with center $O$ and a point $P$ located outside the circle.

Tools: Compass, Straightedge.

Goal: Construct the two tangent lines from point $P$ to the circle.

Steps:

1. Join Center to External Point: Use the straightedge to draw the line segment $OP$ connecting the center $O$ of the circle to the external point $P$.
2. Find the Midpoint of OP: Construct the perpendicular bisector of the line segment $OP$ (I3). Let the midpoint of $OP$ be $M$.
3. Draw Auxiliary Circle: With $M$ as the center and compass radius equal to the length of $MO$ (or $MP$, since $M$ is the midpoint), draw a circle (or a semicircle if drawing space is limited, ensuring it crosses the original circle). This auxiliary circle will pass through both $O$ and $P$ and will intersect the original given circle at two distinct points.
4. Identify the Points of Tangency: Label the two points where the auxiliary circle intersects the original given circle as $T_1$ and $T_2$. These points are the points of tangency for the tangents from $P$.
5. Draw the Tangent Lines: Use the straightedge to draw the line segment connecting the external point $P$ to the point $T_1$. Extend this segment to form the tangent line $PT_1$. Similarly, draw the line segment connecting $P$ to $T_2$ and extend it to form the tangent line $PT_2$.

The lines $PT_1$ and $PT_2$ are the required tangents from point $P$ to the circle.

Justification

Given: Circle center $O$, external point $P$. Points $T_1, T_2$ are intersections of the original circle and the auxiliary circle with diameter $OP$. Lines $PT_1, PT_2$ are drawn.

To Prove: $PT_1$ and $PT_2$ are tangents to the circle at $T_1$ and $T_2$ respectively.

Proof:

Join $OT_1$ and $OT_2$. These are radii of the original circle. Also join $OT_1$ and $OT_2$ in relation to the auxiliary circle.

Consider the auxiliary circle with center $M$ and diameter $OP$. The points $T_1$ and $T_2$ lie on the circumference of this auxiliary circle by construction.

Consider the angle $\angle OT_1P$. This angle is formed by joining a point $T_1$ on the auxiliary circle to the endpoints of its diameter $OP$. An angle inscribed in a semicircle is a right angle. Therefore, the angle $\angle OT_1P$ is a right angle.

This means that the line segment $PT_1$ is perpendicular to the radius $OT_1$ of the original circle at the point $T_1$, which is a point on the original circle.

According to the theorem, "A line perpendicular to the radius at its endpoint on the circle is tangent to the circle", the line $PT_1$ must be a tangent to the original circle at point $T_1$.

Similarly, consider the angle $\angle OT_2P$. This angle is also inscribed in a semicircle of the auxiliary circle (with diameter $OP$). Thus, $\angle OT_2P = 90^\circ$.

This means that the line segment $PT_2$ is perpendicular to the radius $OT_2$ of the original circle at the point $T_2$ on the original circle. Therefore, the line $PT_2$ must be a tangent to the original circle at point $T_2$.

Thus, $PT_1$ and $PT_2$ are the two tangents to the circle from the external point $P$. Q.E.D.

Example

Answer:

Construction of a Pair of Tangents to a Circle with a specific angle between them (using properties of quadrilaterals and tangents)

This construction addresses the problem of drawing two tangents from an external point such that the angle formed by the intersection of these tangents has a specific given measure.

The method relies on the geometric relationship between the angle between the two tangents from an external point and the angle subtended by the segment joining the points of tangency at the center of the circle.

Property Used

Consider a circle with center $O$. Let $P$ be an external point from which two tangents $PT_1$ and $PT_2$ are drawn to the circle, with $T_1$ and $T_2$ being the points of tangency. Join $OT_1$, $OT_2$, $PT_1$, and $PT_2$. This forms the quadrilateral $OT_1PT_2$.

From the radius-tangent property, we know that the radius is perpendicular to the tangent at the point of tangency:

$\angle OT_1P = 90^\circ$

(Radius $OT_1 \perp$ tangent $PT_1$)

$\angle OT_2P = 90^\circ$

(Radius $OT_2 \perp$ tangent $PT_2$)

The sum of the interior angles in any quadrilateral is $360^\circ$. In quadrilateral $OT_1PT_2$:

Substitute the known right angles: Subtracting $180^\circ$ from both sides: This equation shows that the angle between the radii at the center ($\angle T_1OT_2$) and the angle between the tangents at the external point ($\angle T_1PT_2$) are supplementary. Their sum is $180^\circ$.

Therefore, if the desired angle between the tangents is $\theta$, the required angle between the radii joining the center to the points of tangency must be $180^\circ - \theta$. This relationship is the key to the construction.

Construction Steps

Given: A circle with center $O$ and radius $r$, and the desired measure of the angle between the two tangents from an external point, $\theta$.

Tools: Ruler, Compass, Straightedge, Protractor (needed for constructing the required central angle, unless it's a standard constructible angle).

Goal: Construct a pair of tangents $PT_1$ and $PT_2$ to the circle such that $\angle T_1PT_2 = \theta$.

Steps:

1. Calculate the Central Angle: Determine the required angle between the radii at the center using the supplementary property:

   Required Central Angle ($\alpha$) = $180^\circ - \theta$

   ... (iv)

   Ensure $0^\circ < \theta < 180^\circ$ for a valid angle between tangents from an external point. This implies $0^\circ < \alpha < 180^\circ$.
2. Draw First Radius: Draw any radius of the circle. Let this be the radius $OT_1$. $T_1$ will be one of the points of tangency.
3. Construct the Central Angle: At the center $O$, construct an angle equal to the calculated central angle $\alpha$, using $OT_1$ as one arm. Draw the second arm of this angle, say ray $OR$, starting from $O$. This ray $OR$ intersects the circle at the second point of tangency, $T_2$. So, $\angle T_1OT_2 = \alpha = 180^\circ - \theta$.
4. Construct Perpendiculars (Tangents) at Points of Tangency: The tangents at $T_1$ and $T_2$ must be perpendicular to the radii $OT_1$ and $OT_2$ respectively.
   • At point $T_1$ on the circle, construct a line perpendicular to the radius $OT_1$ using the method described in I1 (Tangent at a point on the circle). This line is the first tangent.
   • At point $T_2$ on the circle, construct a line perpendicular to the radius $OT_2$ using the same method. This line is the second tangent.
5. Locate the External Point: The two tangent lines constructed in Step 4 will intersect at an external point. Extend these two tangent lines until they meet. Label the point of intersection as $P$.

The lines $PT_1$ and $PT_2$ are the required pair of tangents to the circle, and the angle between them, $\angle T_1PT_2$, is equal to the given angle $\theta$.

Justification:

By construction, $OT_1 \perp PT_1$ and $OT_2 \perp PT_2$ because the tangents are constructed as lines perpendicular to the radii at the points on the circle. Thus, $PT_1$ and $PT_2$ are indeed tangent lines to the circle at $T_1$ and $T_2$.

Let $P$ be the intersection point of the tangents. Consider the quadrilateral $OT_1PT_2$. We have:

$\angle OT_1P = 90^\circ$

... (v)

$\angle OT_2P = 90^\circ$

... (vi)

$\angle T_1OT_2 = 180^\circ - \theta$

(By construction, Step 3)

The sum of angles in $OT_1PT_2$ is $360^\circ$. Substitute the known angles into (vii): Subtract $360^\circ$ from both sides: Thus, the angle between the tangents at point $P$ is equal to the given angle $\theta$. Q.E.D.

Example

Answer:

Justification of Tangent Constructions

The justifications for the constructions of tangents to a circle provide the mathematical reasoning why the steps taken produce the desired tangent lines. These justifications rely on fundamental theorems and properties of circles, angles, and perpendicular lines.

Justification for Tangent at a Point P on the Circle (Subheading I1)

Construction Performed: Given a circle with center $O$ and point $P$ on it, we drew the radius $OP$ and constructed a line $l$ through $P$ such that $l$ is perpendicular to $OP$.

Theorem Used for Justification:

The primary theorem used is: A line in the plane of a circle is tangent to the circle if and only if it is perpendicular to a radius at its endpoint on the circle.

This is a two-way theorem (an "if and only if" statement). One direction states: "If a line is perpendicular to a radius at its endpoint on the circle, then the line is tangent to the circle."

Justification Steps:

1. The point $P$ is on the circle by construction.
2. $OP$ is the radius drawn to the point $P$.
3. The line $l$ is constructed such that it passes through $P$ and $l \perp OP$.
4. Applying the cited theorem, since line $l$ passes through $P$ (an endpoint of radius $OP$ on the circle) and is perpendicular to $OP$ at $P$, the line $l$ must be the tangent to the circle at $P$. Q.E.D.

Justification for Tangents from an External Point P (Subheading I2)

Construction Performed: Given a circle with center $O$ and external point $P$, we drew $OP$, found its midpoint $M$, drew an auxiliary circle with center $M$ and radius $MO$ intersecting the original circle at $T_1$ and $T_2$, and drew lines $PT_1$ and $PT_2$.

Theorems Used for Justification:

• Angle in a Semicircle Theorem: The angle subtended by a diameter at any point on the circumference of a circle is a right angle ($90^\circ$).
• Tangent Property Theorem: A line is tangent to a circle if and only if it is perpendicular to the radius at the point of contact.

Justification Steps:

1. The auxiliary circle with center $M$ has $OP$ as its diameter (since $M$ is the midpoint of $OP$ and the radius is $MO=MP$).
2. Points $T_1$ and $T_2$ lie on the circumference of this auxiliary circle (by construction).
3. Consider $\triangle OT_1P$. Point $T_1$ is on the circumference of the auxiliary circle, and $OP$ is the diameter. Therefore, the angle $\angle OT_1P$, which subtends the diameter $OP$ in the auxiliary circle, is an angle in a semicircle. By the Angle in a Semicircle Theorem, $\angle OT_1P = 90^\circ$.
4. This means the line $PT_1$ is perpendicular to the segment $OT_1$. Since $OT_1$ is a radius of the original circle and $T_1$ is a point on the original circle, this means $PT_1$ is perpendicular to the radius $OT_1$ at its endpoint $T_1$ on the circle.
5. By the Tangent Property Theorem, a line perpendicular to a radius at its endpoint on the circle is tangent to the circle. Therefore, $PT_1$ is tangent to the original circle at $T_1$.
6. A similar argument applies to $\triangle OT_2P$. $\angle OT_2P$ is also an angle in the semicircle of the auxiliary circle with diameter $OP$. Hence, $\angle OT_2P = 90^\circ$.
7. Thus, $PT_2$ is perpendicular to the radius $OT_2$ at its endpoint $T_2$ on the original circle. By the Tangent Property Theorem, $PT_2$ is tangent to the original circle at $T_2$.

Therefore, $PT_1$ and $PT_2$ are the required tangents from external point $P$. Q.E.D.

Justification for Tangents with a Given Angle Between Them (Subheading I3)

Construction Performed: Given a circle with center $O$ and desired tangent angle $\theta$, we calculated central angle $\alpha = 180^\circ - \theta$. We drew radii $OT_1$ and $OT_2$ such that $\angle T_1OT_2 = \alpha$. We then constructed lines perpendicular to $OT_1$ at $T_1$ and to $OT_2$ at $T_2$, and labeled their intersection point $P$.

Theorems Used for Justification:

• Tangent Property Theorem: A line is tangent to a circle if and only if it is perpendicular to the radius at the point of contact.
• Quadrilateral Angle Sum Property: The sum of the interior angles of any quadrilateral is $360^\circ$.

Justification Steps:

1. By construction, the line drawn through $T_1$ is perpendicular to the radius $OT_1$ at point $T_1$ on the circle. By the Tangent Property Theorem, this line is tangent to the circle at $T_1$.
2. Similarly, the line drawn through $T_2$ is perpendicular to the radius $OT_2$ at point $T_2$ on the circle. By the Tangent Property Theorem, this line is tangent to the circle at $T_2$.
3. These two tangent lines intersect at point $P$. Consider the quadrilateral $OT_1PT_2$. The vertices are the center $O$, the two points of tangency $T_1$ and $T_2$, and the external intersection point $P$.
4. In quadrilateral $OT_1PT_2$, we know the following angles:

   $\angle OT_1P = 90^\circ$

   (Radius $\perp$ Tangent at $T_1$)

   $\angle OT_2P = 90^\circ$

   (Radius $\perp$ Tangent at $T_2$)

   $\angle T_1OT_2 = 180^\circ - \theta$

   (By construction, Step 3)

5. The sum of the interior angles of quadrilateral $OT_1PT_2$ must be $360^\circ$.

   $\angle T_1PT_2 + \angle OT_1P + \angle T_1OT_2 + \angle OT_2P = 360^\circ$

   ... (xii)

6. Substitute the known values into equation (xii):

   $\angle T_1PT_2 + 90^\circ + (180^\circ - \theta) + 90^\circ = 360^\circ$

   ... (xiii)

   $\angle T_1PT_2 + 360^\circ - \theta = 360^\circ$

   ... (xiv)

7. Subtract $360^\circ$ from both sides of equation (xiv):

   $\angle T_1PT_2 - \theta = 0$

   ... (xv)

   $\angle T_1PT_2 = \theta$

   ... (xvi)

8. Thus, the angle formed by the constructed tangents $PT_1$ and $PT_2$ at their intersection point $P$ is exactly equal to the given angle $\theta$.

These justifications, based on proven geometric theorems, confirm that the steps for tangent constructions are valid and achieve the desired outcomes. Q.E.D.