krit.club logo

Geometry and Trigonometry - Tangents to a circle and their properties-extended

Grade 9IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

A tangent to a circle is a straight line that touches the circle at exactly one point, called the point of tangency. The radius of the circle drawn to this point is always perpendicular (90∘90^\circ) to the tangent line.

•

Two tangents drawn to a circle from the same external point are equal in length. This creates a kite shape with the radii, where the line connecting the external point to the center bisects the angle between the tangents.

Diagram showing two tangents PA and PB from point P to a circle with center O.
•

The angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment. This is known as the Alternate Segment Theorem.

•

If two circles touch each other externally, the distance between their centers is equal to the sum of their radii (r1+r2r_1 + r_2). If they touch internally, the distance is the difference of their radii (r1−r2r_1 - r_2).

📐Formulae

In △OPT, where ∠T=90∘:OP2=OT2+PT2\text{In } \triangle OPT, \text{ where } \angle T = 90^\circ: OP^2 = OT^2 + PT^2

Length of tangent (t)=d2−r2 where d is the distance from the center and r is the radius.\text{Length of tangent } (t) = \sqrt{d^2 - r^2} \text{ where } d \text{ is the distance from the center and } r \text{ is the radius.}

∠AOB+∠APB=180∘ (for tangents from point P to points A,B and center O)\angle AOB + \angle APB = 180^\circ \text{ (for tangents from point } P \text{ to points } A, B \text{ and center } O\text{)}

Angle in alternate segment: ∠BAT=∠BCA\text{Angle in alternate segment: } \angle BAT = \angle BCA

💡Examples

Problem 1:

A tangent PAPA is drawn from an external point PP to a circle with center OO and radius 7 cm7\text{ cm}. If the length of the tangent PAPA is 24 cm24\text{ cm}, calculate the distance of point PP from the center OO.

Solution:

  1. In △OAP\triangle OAP, the radius OAOA is perpendicular to the tangent PAPA, so ∠OAP=90∘\angle OAP = 90^\circ.
  2. Use the Pythagorean theorem: OP2=OA2+PA2OP^2 = OA^2 + PA^2.
  3. Substitute the values: OP2=72+242OP^2 = 7^2 + 24^2.
  4. OP2=49+576=625OP^2 = 49 + 576 = 625.
  5. OP=625=25 cmOP = \sqrt{625} = 25\text{ cm}.

Explanation:

Since the radius is perpendicular to the tangent at the point of contact, we form a right-angled triangle. We then apply the Pythagorean theorem to find the hypotenuse OPOP.

Problem 2:

Two tangents PQPQ and PRPR are drawn to a circle with center OO from an external point PP. If ∠QPR=50∘\angle QPR = 50^\circ, find the measure of ∠OQR\angle OQR.

Solution:

  1. In quadrilateral OQPROQPR, ∠OQP=90∘\angle OQP = 90^\circ and ∠ORP=90∘\angle ORP = 90^\circ (tangent-radius property).
  2. The sum of angles in a quadrilateral is 360∘360^\circ, so ∠QOR=360∘−(90∘+90∘+50∘)=130∘\angle QOR = 360^\circ - (90^\circ + 90^\circ + 50^\circ) = 130^\circ.
  3. In △OQR\triangle OQR, OQ=OROQ = OR (radii of the same circle), making it an isosceles triangle.
  4. Therefore, ∠OQR=∠ORQ\angle OQR = \angle ORQ.
  5. ∠OQR=180∘−130∘2=50∘2=25∘\angle OQR = \frac{180^\circ - 130^\circ}{2} = \frac{50^\circ}{2} = 25^\circ.

Explanation:

First, we find the central angle ∠QOR\angle QOR using the fact that the angles of a quadrilateral sum to 360∘360^\circ. Then, we use the properties of an isosceles triangle formed by the two radii to find the base angle.

Problem 3:

Find the value of xx in the following subtraction problem involving distances from a tangent point: 150−86x\begin{array}{r} 150 \\ - 86 \\ \hline x \end{array}

Solution:

150−8664\begin{array}{r} 150 \\ - 86 \\ \hline 64 \end{array} So, x=64x = 64.

Explanation:

This is a simple vertical arithmetic subtraction to determine a remaining segment length on a line tangent to a circle.

Problem 4:

In the given figure, PTPT is a tangent to a circle with center OO at point TT. If OT=5 cmOT = 5\text{ cm} and OP=13 cmOP = 13\text{ cm}, find the length of the tangent segment PTPT.

Right triangle OTP with radius 5, hypotenuse 13, and tangent PT.

Solution:

  1. Since PTPT is a tangent at TT, ∠OTP=90∘\angle OTP = 90^\circ according to the tangent-radius theorem.
  2. △OTP\triangle OTP is a right-angled triangle. By Pythagoras' theorem: OT2+PT2=OP2OT^2 + PT^2 = OP^2
  3. Substitute the known values: 52+PT2=1325^2 + PT^2 = 13^2 25+PT2=16925 + PT^2 = 169
  4. Solve for PTPT: PT2=169−25PT^2 = 169 - 25 PT2=144PT^2 = 144 PT=144=12PT = \sqrt{144} = 12 Therefore, the length of the tangent PTPT is 12 cm12\text{ cm}.

Explanation:

The radius drawn to the point of tangency is always perpendicular to the tangent line, allowing the use of the Pythagorean theorem to find missing side lengths in the resulting right triangle.

Problem 5:

Two tangents PAPA and PBPB are drawn to a circle with center OO from an external point PP. If the angle between the tangents ∠APB=70∘\angle APB = 70^\circ, find the measure of the central angle ∠AOB\angle AOB.

Quadrilateral OAPB formed by two tangents and two radii.

Solution:

  1. In the quadrilateral OAPBOAPB, ∠OAP=90∘\angle OAP = 90^\circ and ∠OBP=90∘\angle OBP = 90^\circ because radii are perpendicular to tangents at the point of contact.
  2. The sum of angles in a quadrilateral is 360∘360^\circ: ∠AOB+∠OAP+∠APB+∠OBP=360∘\angle AOB + \angle OAP + \angle APB + \angle OBP = 360^\circ
  3. Substitute the known values: ∠AOB+90∘+70∘+90∘=360∘\angle AOB + 90^\circ + 70^\circ + 90^\circ = 360^\circ ∠AOB+250∘=360∘\angle AOB + 250^\circ = 360^\circ
  4. Calculate ∠AOB\angle AOB: ∠AOB=360∘−250∘=110∘\angle AOB = 360^\circ - 250^\circ = 110^\circ Therefore, ∠AOB=110∘\angle AOB = 110^\circ.

Explanation:

Because the angles at the points of tangency are both 90∘90^\circ, the angle between the tangents and the angle at the center are supplementary (they add up to 180∘180^\circ).