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Circles - Tangent to 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 straight line that intersects the circle at exactly one point, known as the point of contact. Visually, imagine a line just 'touching' the edge of a circular boundary without crossing into the interior.

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The tangent at any point of a circle is perpendicular to the radius through the point of contact. This means if you draw a line from the center OO to the point of contact PP, the angle between the radius OPOP and the tangent line is always 90∘90^{\circ}.

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From a point inside a circle, no tangent can be drawn. From a point on the circle, exactly one tangent can be drawn. From a point outside the circle, exactly two tangents can be drawn to the circle.

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The lengths of tangents drawn from an external point to a circle are equal. Visually, if you pick a point PP outside a circle and draw two lines PAPA and PBPB that touch the circle at AA and BB, the segment PAPA will be equal in length to PBPB.

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The center of the circle lies on the angle bisector of the angle between the two tangents drawn from an external point. This creates a symmetric visual where the line segment connecting the center OO to the external point PP divides the angle ∠APB\angle APB into two equal parts.

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The quadrilateral formed by the center of the circle, the two points of contact, and the external point is a cyclic quadrilateral because the sum of the opposite angles (the two 90∘90^{\circ} angles at the points of contact) is 180∘180^{\circ}. Consequently, the angle between the two tangents and the angle subtended by the radii at the center are supplementary: ∠APB+∠AOB=180∘\angle APB + \angle AOB = 180^{\circ}.

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A secant is a line that intersects a circle in two points. In contrast, the tangent is a limiting case of the secant when the two endpoints of the corresponding chord coincide.

📐Formulae

Length of tangent segment from external point PP to contact point AA: PA=OP2−r2PA = \sqrt{OP^2 - r^2} (where OO is center and rr is radius)

Pythagoras Theorem in △OAP\triangle OAP: OP2=OA2+PA2OP^2 = OA^2 + PA^2

Angle relation: ∠APB+∠AOB=180∘\angle APB + \angle AOB = 180^{\circ} (where A,BA, B are points of contact and PP is the external point)

In △TPQ\triangle TPQ (where TT is external point and P,QP, Q are contact points): ∠TPQ=∠TQP=12(180∘−∠PTQ)\angle TPQ = \angle TQP = \frac{1}{2}(180^{\circ} - \angle PTQ)

Area of quadrilateral OAPB=2×(12×r×PA)=r×PAOAPB = 2 \times (\frac{1}{2} \times r \times PA) = r \times PA

💡Examples

Problem 1:

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

Solution:

  1. Let OO be the center of the circle and PP be the point of contact.
  2. In △OPQ\triangle OPQ, the radius OPOP is perpendicular to the tangent PQPQ. Therefore, △OPQ\triangle OPQ is a right-angled triangle at PP.
  3. Using the Pythagoras Theorem: OQ2=OP2+PQ2OQ^2 = OP^2 + PQ^2
  4. Substitute the given values: 252=OP2+24225^2 = OP^2 + 24^2
  5. 625=OP2+576625 = OP^2 + 576
  6. OP2=625−576=49OP^2 = 625 - 576 = 49
  7. OP=49=7OP = \sqrt{49} = 7 cm.

Explanation:

This problem uses the fundamental 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 TPTP and TQTQ are drawn to a circle with center OO from an external point TT. Prove that ∠PTQ=2∠OPQ\angle PTQ = 2\angle OPQ.

Solution:

  1. Let ∠PTQ=θ\angle PTQ = \theta.
  2. Since lengths of tangents from an external point are equal, TP=TQTP = TQ. Thus, △TPQ\triangle TPQ is an isosceles triangle.
  3. In △TPQ\triangle TPQ, ∠TPQ=∠TQP=12(180∘−θ)=90∘−θ2\angle TPQ = \angle TQP = \frac{1}{2}(180^{\circ} - \theta) = 90^{\circ} - \frac{\theta}{2}.
  4. We know the radius OP⊥TPOP \perp TP, so ∠OPT=90∘\angle OPT = 90^{\circ}.
  5. ∠OPQ=∠OPT−∠TPQ\angle OPQ = \angle OPT - \angle TPQ
  6. ∠OPQ=90∘−(90∘−θ2)=θ2\angle OPQ = 90^{\circ} - (90^{\circ} - \frac{\theta}{2}) = \frac{\theta}{2}.
  7. Therefore, θ=2∠OPQ\theta = 2\angle OPQ, which means ∠PTQ=2∠OPQ\angle PTQ = 2\angle OPQ.

Explanation:

This proof relies on the properties of isosceles triangles formed by tangents and the 90∘90^{\circ} angle relationship between the radius and the tangent.