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Circles - Prove tangent at point of contact is perpendicular to radius

Grade 10CBSE

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

🔑Concepts

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The fundamental theorem of circles states that the tangent at any point of a circle is perpendicular to the radius through the point of contact. This means if XYXY is a tangent to a circle with center OO at point PP, then OP⊥XYOP \perp XY, forming a 90∘90^{\circ} angle.

Circle with radius OP perpendicular to tangent XY at point P
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The proof of this theorem uses the fact that the shortest distance from a point to a line is the perpendicular distance. For any point QQ on the tangent (other than PP), OQOQ must be greater than OPOP because QQ lies outside the circle. Since OPOP is the shortest distance from OO to line XYXY, OP⊥XYOP \perp XY.

Geometry showing point Q on tangent outside circle making OQ longer than OP
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Tangent segments from an external point to a circle are equal in length. If PAPA and PBPB are tangents from PP to a circle with center OO, then PA=PBPA = PB. This results in △OAP\triangle OAP being congruent to △OBP\triangle OBP by the RHS congruence rule.

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The line joining the center to the external point PP bisects the angle between the two tangents (∠APB\angle APB) and also bisects the angle between the radii at the center (∠AOB\angle AOB).

📐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.

Problem 3:

In the given figure, XYXY and X′Y′X'Y' are two parallel tangents to a circle with center OO and another tangent ABAB with point of contact CC intersecting XYXY at AA and X′Y′X'Y' at BB. Prove that ∠AOB=90∘\angle AOB = 90^{\circ}.

Circle with two parallel tangents and a third intersecting tangent forming a 90 degree angle at center

Solution:

  1. Connect OCOC. In △OPA\triangle OPA and △OCA\triangle OCA: OP=OCOP = OC (radii) AP=ACAP = AC (tangents from AA) OA=OAOA = OA (common) ∴△OPA≅△OCA\therefore \triangle OPA \cong \triangle OCA (SSS rule), which means ∠POA=∠COA\angle POA = \angle COA.
  2. Similarly, △OQB≅△OCB\triangle OQB \cong \triangle OCB, which means ∠QOB=∠COB\angle QOB = \angle COB.
  3. Since POQPOQ is a diameter (straight line), ∠POA+∠COA+∠COB+∠QOB=180∘\angle POA + \angle COA + \angle COB + \angle QOB = 180^{\circ}.
  4. 2∠COA+2∠COB=180∘  ⟹  2(∠COA+∠COB)=180∘2\angle COA + 2\angle COB = 180^{\circ} \implies 2(\angle COA + \angle COB) = 180^{\circ}.
  5. ∠AOB=180∘2=90∘\angle AOB = \frac{180^{\circ}}{2} = 90^{\circ}.

Explanation:

By proving the congruence of the triangles formed by the radii and tangents, we establish that the center line bisects the angles. The sum of angles on a diameter line leads to the 90∘90^{\circ} conclusion.

Problem 4:

A tangent PQPQ at a point PP of a circle of radius 55 cm meets a line through the center OO at a point QQ so that OQ=12OQ = 12 cm. Find the length PQPQ.

Right angled triangle OPQ with OP as radius and PQ as tangent

Solution:

  1. According to the theorem, the radius OPOP is perpendicular to the tangent PQPQ at the point of contact PP.
  2. Therefore, △OPQ\triangle OPQ is a right-angled triangle with ∠OPQ=90∘\angle OPQ = 90^{\circ}.
  3. Using Pythagoras Theorem: OQ2=OP2+PQ2OQ^2 = OP^2 + PQ^2.
  4. 122=52+PQ212^2 = 5^2 + PQ^2
  5. 144=25+PQ2144 = 25 + PQ^2
  6. PQ2=144−25=119PQ^2 = 144 - 25 = 119
  7. PQ=119PQ = \sqrt{119} cm.

Explanation:

We apply the perpendicular property of the radius to the tangent to create a right triangle and then use the Pythagoras Theorem to solve for the missing side.