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Circles - Properties of Tangents

Grade 10CBSE

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

🔑Concepts

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Definition of a Tangent: A tangent to a circle is a straight line that intersects the circle at exactly one point, known as the point of contact. Visually, imagine a line that just grazes the edge of a circle without entering its interior. The distance from the center of the circle to this line is always equal to the radius rr.

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Theorem 1 (Perpendicularity Property): The tangent at any point of a circle is perpendicular to the radius through the point of contact. If we draw a radius from the center OO to the point of contact PP, then the angle between the radius OPOP and the tangent line is always 90∘90^{\circ}.

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Tangents from an External Point: From any point outside a circle, exactly two tangents can be drawn to the circle. If we denote the external point as PP and the points where the tangents touch the circle as AA and BB, the line segments PAPA and PBPB are called the lengths of the tangents. No tangents can be drawn from a point inside the circle.

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Theorem 2 (Equality of Tangents): The lengths of tangents drawn from an external point to a circle are equal. This means PA=PBPA = PB. Visually, these two tangents and the chord joining the points of contact form an isosceles triangle △PAB\triangle PAB.

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Properties of the Tangent Quadrilateral: When two tangents PAPA and PBPB are drawn from an external point PP to a circle with center OO, the quadrilateral OAPBOAPB is formed. Since ∠OAP=90∘\angle OAP = 90^{\circ} and ∠OBP=90∘\angle OBP = 90^{\circ}, the sum of the other two angles ∠APB+∠AOB=180∘\angle APB + \angle AOB = 180^{\circ}. This makes OAPBOAPB a cyclic quadrilateral.

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Center Bisector Property: The line segment joining the center OO to the external point PP bisects the angle between the two tangents (∠APB\angle APB) and also bisects the angle between the two radii (∠AOB\angle AOB). Visually, the line OPOP acts as an axis of symmetry for the two tangents.

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Concept of Secant vs Tangent: While a tangent touches the circle at only one point, a secant is a line that intersects the circle at two distinct points. A tangent can be viewed as a limiting case of a secant when the two intersection points coincide.

📐Formulae

Length of tangent L=d2−r2L = \sqrt{d^2 - r^2}, where dd is the distance from the external point to the center and rr is the radius.

Pythagorean relation in △OTP\triangle OTP: OP2=OT2+PT2OP^2 = OT^2 + PT^2, where OTOT is the radius (rr) and PTPT is the tangent length.

Angle Supplementary Property: ∠AOB+∠APB=180∘\angle AOB + \angle APB = 180^{\circ}.

Perpendicularity: Radius⊥Tangent⇒∠OPT=90∘Radius \perp Tangent \Rightarrow \angle OPT = 90^{\circ}.

Equality of lengths: PA=PBPA = PB (for tangents from external point PP).

💡Examples

Problem 1:

A point PP is 13 cm13\text{ cm} away from the center of a circle. If the length of the tangent drawn from PP to the circle is 12 cm12\text{ cm}, find the radius of the circle.

Solution:

  1. Let OO be the center of the circle and TT be the point of contact.
  2. In △OTP\triangle OTP, the radius OTOT is perpendicular to the tangent PTPT, so ∠OTP=90∘\angle OTP = 90^{\circ}.
  3. Using the Pythagorean Theorem: OP2=OT2+PT2OP^2 = OT^2 + PT^2.
  4. Substitute the given values: 132=r2+12213^2 = r^2 + 12^2.
  5. 169=r2+144169 = r^2 + 144.
  6. r2=169−144=25r^2 = 169 - 144 = 25.
  7. r=25=5 cmr = \sqrt{25} = 5\text{ cm}.

Explanation:

This problem uses the property that the radius is perpendicular to the tangent at the point of contact, creating a right-angled triangle where the distance from the center is the hypotenuse.

Problem 2:

Two tangents PAPA and PBPB are drawn to a circle with center OO from an external point PP. If ∠APB=80∘\angle APB = 80^{\circ}, calculate the value of ∠POA\angle POA.

Solution:

  1. We know that the line joining the external point PP to the center OO bisects the angle between the tangents.
  2. Therefore, ∠APO=12∠APB=12×80∘=40∘\angle APO = \frac{1}{2} \angle APB = \frac{1}{2} \times 80^{\circ} = 40^{\circ}.
  3. In △OAP\triangle OAP, we know ∠OAP=90∘\angle OAP = 90^{\circ} (radius perpendicular to tangent).
  4. The sum of angles in △OAP\triangle OAP is 180∘180^{\circ}.
  5. ∠POA+∠OAP+∠APO=180∘\angle POA + \angle OAP + \angle APO = 180^{\circ}.
  6. ∠POA+90∘+40∘=180∘\angle POA + 90^{\circ} + 40^{\circ} = 180^{\circ}.
  7. ∠POA=180∘−130∘=50∘\angle POA = 180^{\circ} - 130^{\circ} = 50^{\circ}.

Explanation:

This solution relies on two properties: first, that the line from the center to the external point bisects the angle between the tangents; and second, the angle sum property of the right-angled triangle formed by the radius and tangent.