Properties of Tangents to a Circle

In geometry, a tangent is a line that touches a curve at a single point. For circles, a tangent line touches the circle at exactly one point. The properties of tangents are fundamental in understanding the relationships between circles and lines. This concept specifically deals with tangents drawn from a point outside the circle.

Formulae

The key takeaway from this concept is the following:

• If two tangents are drawn from an external point to a circle, then the lengths of these tangents are equal.

Examples

Let’s look at some examples:

Example-1: A circle has center O. Point P lies outside the circle. PA and PB are tangents drawn from P to the circle. If PA = 5 cm, then PB = ?

Solution: Since PA and PB are tangents from the same external point P, according to the theorem, they are equal in length. Therefore, PB = 5 cm.

Example-2: In the given figure, PT and PS are tangents to the circle with center O. If $\angle TPS = 60^\circ$ and the radius of the circle is 4 cm, find the length of TP. (Assume the angle at the center subtended by the tangent is 90 degrees)

Solution: Since PT and PS are tangents from the same external point P, PT = PS. Therefore, $\triangle TPS$ is an isosceles triangle. So, $\angle PTS = \angle PST = \frac{180^\circ – 60^\circ}{2} = 60^\circ$. This means $\triangle TPS$ is an equilateral triangle. Hence, TP = PS = TS. Since $\angle OTP = 90^\circ$, we need to establish the exact relationship of the lengths. We can use trigonometric ratios for $\triangle OTP$. We know that $\angle TPO = 30^\circ$ since it’s half of $\angle TPS$. Therefore, using the tangent function, $$\tan(\angle TPO) = \frac{OT}{TP}$$ $$\tan(30^\circ) = \frac{4}{TP}$$ $$\frac{1}{\sqrt{3}} = \frac{4}{TP}$$ Therefore, TP = $4\sqrt{3}$ cm.

Theorem with Proof

Theorem: The lengths of the tangents drawn from an external point to a circle are equal.

Proof:

1. Given: A circle with center O. P is a point outside the circle. PA and PB are tangents to the circle from P, where A and B are the points of tangency.
2. To Prove: PA = PB
3. Construction: Join OA, OB, and OP.
4. Proof:
   • $\angle OAP = 90^\circ$ and $\angle OBP = 90^\circ$ (Tangent is perpendicular to the radius at the point of contact)
   • In $\triangle OAP$ and $\triangle OBP$:
   • OA = OB (Radii of the same circle)
   • OP = OP (Common side)
   • $\angle OAP = \angle OBP = 90^\circ$
   • Therefore, $\triangle OAP \cong \triangle OBP$ (By RHS congruence criterion)
   • Hence, PA = PB (Corresponding sides of congruent triangles are equal)

Common mistakes by students

Students often make these mistakes:

• Forgetting the Theorem: Students may not remember that the lengths of tangents from an external point are equal, leading them to incorrectly assume different lengths.
• Misinterpreting the Figure: Students may fail to recognize the right angles formed by the radius and the tangent at the point of contact, and that the triangle formed is a right angle triangle.
• Incorrect Application of Other Theorems: Applying the wrong theorems or formulas when solving problems involving tangents and circles. For example, incorrectly using Pythagoras theorem or trigonometric ratios.

Real Life Application

The properties of tangents are applicable in various real-life scenarios such as:

• Designing circular tracks: When designing a circular running track or a road, engineers need to ensure that the tangents at any point on the track are smooth and continuous.
• Engineering projects: In various engineering projects that involve curves and circles, such as designing bridges or tunnels, the properties of tangents are essential for calculations.
• Navigation: Calculating the shortest distance to a circular destination, like a well or silo, where the point of tangency would define the shortest route (perpendicular distance from center).

Fun Fact

Did you know that the concept of tangents is crucial in the design of gears? The shape of the teeth on gears is often based on tangent lines to ensure smooth and efficient power transmission.

Recommended YouTube Videos for Deeper Understanding

Q.1 A point P is 10 cm away from the center of a circle of radius 6 cm. Two tangents PA and PB are drawn to the circle. What is the length of each tangent?

Check Solution

Ans: B

Using the Pythagorean theorem in triangle OAP, where O is the center, $OA^2 + PA^2 = OP^2$. Thus, $6^2 + PA^2 = 10^2$. So, $PA^2 = 100 – 36 = 64$, and $PA = 8$ cm.

Q.2 Two tangents are drawn from a point Q to a circle. The length of one tangent is 12 cm. What is the length of the other tangent?

Check Solution

Ans: B

According to the property of tangents, lengths of tangents from an external point are equal.

Q.3 In the given figure, tangents PA and PB are drawn to the circle from an external point P. If the radius of the circle is 5 cm and OP is 13 cm, then find the length of the tangent PA.

Check Solution

Ans: B

Using the Pythagorean theorem in triangle OAP, $OA^2 + PA^2 = OP^2$. Therefore, $5^2 + PA^2 = 13^2$. Hence, $PA^2 = 169 – 25 = 144$, and $PA = 12$ cm.

Q.4 A circle is inscribed in a quadrilateral ABCD, touching sides AB, BC, CD, and DA at P, Q, R, and S respectively. If AB = 10 cm, BC = 12 cm, and AS = 4 cm, what is the length of CD?

Check Solution

Ans: D

Let $AP = AS = 4$ cm. Then, $PB = AB – AP = 10 – 4 = 6$ cm. Since $PB = BQ = 6$ cm. Then, $CQ = BC – BQ = 12 – 6 = 6$ cm. Thus, $CR = CQ = 6$ cm. Also $DR = DS$. We know $CD = CR + RD = 6+DS$. Since the lengths of tangents from a point to a circle are equal, $DS = DA – AS$. Assuming, $DA = DS + AS = 8$ cm, we can find $DS=8-4=4cm$, $CD = CR+RD = 6+DS = 6+4=10$cm

Q.5 Two tangents from a point K to a circle have lengths of $x+3$ cm and $2x-1$ cm. What is the value of $x$?

Check Solution

Ans: C

According to the property of tangents, the lengths of the tangents from an external point to a circle are equal. Therefore, $x+3 = 2x-1$. Solving for x, we get $x = 4$.

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