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Circles - Number of Tangents from a Point on a Circle

Grade 10CBSE

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

🔑Concepts

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A tangent to a circle is a line that intersects the circle at exactly one point, known as the point of contact. Visually, the tangent line touches the outer boundary of the circle without entering its interior.

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The number of tangents depends on the position of the point relative to the circle. If a point lies inside the circle, no tangents can be drawn through it. Visually, any line through an interior point becomes a secant, cutting the circle at two distinct points.

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If a point lies exactly on the circle, there is one and only one tangent passing through that point. Visually, this line is perfectly balanced on the edge of the circle, perpendicular to the radius at that point.

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If a point lies outside the circle, exactly two tangents can be drawn to the circle from that point. Visually, these two lines emerge from the point and 'graze' the circle on opposite sides, meeting the circle at two distinct points of contact.

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The tangent at any point of a circle is perpendicular to the radius through the point of contact. This creates a right angle (∠OPT=90∘\angle OPT = 90^{\circ}) between the radius OTOT and the tangent line PTPT.

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The lengths of the two tangents drawn from an external point to a circle are equal. If PAPA and PBPB are tangents from point PP, then PA=PBPA = PB. Visually, the segments from the external point to the points where they touch the circle form two congruent triangles when connected to the center.

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The center of the circle lies on the angle bisector of the angle between the two tangents. This means the line joining the external point to the center divides the angle between the tangents into two equal parts: ∠APO=∠BPO\angle APO = \angle BPO.

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The angle between the two tangents from an external point and the angle subtended by the line segments joining the points of contact at the center are supplementary. This means ∠APB+∠AOB=180∘\angle APB + \angle AOB = 180^{\circ}.

📐Formulae

PA=PBPA = PB (Equality of tangent lengths from external point PP)

PT=OP2−r2PT = \sqrt{OP^2 - r^2} (Length of tangent PTPT using Pythagoras theorem, where OPOP is distance from center and rr is radius)

∠OPT=90∘\angle OPT = 90^{\circ} (Radius ⊥\perp Tangent at point of contact)

∠APB+∠AOB=180∘\angle APB + \angle AOB = 180^{\circ} (Sum of opposite angles in quadrilateral OAPBOAPB)

sin⁡(∠APO)=rOP\sin(\angle APO) = \frac{r}{OP} (Trigonometric relation in △OAP\triangle OAP)

💡Examples

Problem 1:

A point QQ is at a distance of 2525 cm from the center of a circle and the length of the tangent QTQT to the circle is 2424 cm. Find the radius of the circle.

Solution:

  1. Let OO be the center of the circle and rr be the radius.
  2. In the right-angled triangle △OTQ\triangle OTQ, ∠OTQ=90∘\angle OTQ = 90^{\circ} because the radius is perpendicular to the tangent at the point of contact.
  3. Using the Pythagoras Theorem: OQ2=OT2+QT2OQ^2 = OT^2 + QT^2
  4. Substitute the given values: 252=r2+24225^2 = r^2 + 24^2
  5. 625=r2+576625 = r^2 + 576
  6. r2=625−576=49r^2 = 625 - 576 = 49
  7. r=49=7r = \sqrt{49} = 7 cm.

Explanation:

We identify the right-angled triangle formed by the radius, the tangent, and the line from the center to the external point, then apply the Pythagoras theorem to find the missing side.

Problem 2:

Two tangents PAPA and PBPB are drawn to a circle with center OO from an external point PP. If ∠AOB=110∘\angle AOB = 110^{\circ}, then find the measure of ∠APB\angle APB.

Solution:

  1. In quadrilateral OAPBOAPB, ∠OAP=90∘\angle OAP = 90^{\circ} and ∠OBP=90∘\angle OBP = 90^{\circ} because the radius is perpendicular to the tangent at the point of contact.
  2. The sum of interior angles of a quadrilateral is 360∘360^{\circ}.
  3. Therefore, ∠OAP+∠APB+∠OBP+∠AOB=360∘\angle OAP + \angle APB + \angle OBP + \angle AOB = 360^{\circ}.
  4. Substituting the values: 90∘+∠APB+90∘+110∘=360∘90^{\circ} + \angle APB + 90^{\circ} + 110^{\circ} = 360^{\circ}.
  5. 290∘+∠APB=360∘290^{\circ} + \angle APB = 360^{\circ}.
  6. ∠APB=360∘−290∘=70∘\angle APB = 360^{\circ} - 290^{\circ} = 70^{\circ}.

Explanation:

This problem uses the property that the angles formed by the radii and tangents at the points of contact are 90∘90^{\circ}, and that the angle at the center and the angle between tangents are supplementary.