krit.club logo

Geometry - Constructions (Tangents to a circle, Circumscribing and Inscribing a circle)

Grade 10ICSE

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

🔑Concepts

•

The tangent at any point on a circle is perpendicular to the radius through the point of contact. This fundamental property allows us to construct tangents by drawing a perpendicular to the radius at the point where it meets the circumference.

A circle with center O and a tangent line at point P showing a 90-degree angle with the radius.
•

The circumcircle of a triangle is constructed by finding the intersection of the perpendicular bisectors of any two sides. This point is the circumcenter (CC), which is equidistant from all three vertices (A,B,CA, B, C) of the triangle.

A triangle ABC with its circumcircle passing through all three vertices.
•

The incircle of a triangle is the largest circle that can fit inside the triangle, touching all three sides. It is constructed by finding the intersection of the internal angle bisectors of the triangle.

An incircle inside a triangle where the center is the intersection of angle bisectors.
•

To construct a tangent to a circle from an external point, we join the point to the center, bisect this line segment, and draw a second circle with the midpoint as center and radius equal to half the segment length.

📐Formulae

Angle between Radius and Tangent: ∠OPT=90∘\angle OPT = 90^{\circ}

Length of Tangents from External Point: PA=PBPA = PB

Tangent-Secant Theorem: PT2=PA⋅PBPT^2 = PA \cdot PB (where PTPT is the tangent and PABPAB is a secant line)

Inradius (rr) of a triangle: r=Area of triangle (A)Semi-perimeter (s)r = \frac{\text{Area of triangle (A)}}{\text{Semi-perimeter (s)}}

Circumradius (RR) of a triangle: R=abc4AR = \frac{abc}{4A} (where a,b,ca, b, c are side lengths)

Semi-perimeter: s=a+b+c2s = \frac{a + b + c}{2}

💡Examples

Problem 1:

Construct a triangle ABCABC with AB=6 cmAB = 6 \text{ cm}, BC=7 cmBC = 7 \text{ cm}, and AC=5 cmAC = 5 \text{ cm}. Construct the incircle of this triangle.

Solution:

  1. Draw the base BC=7 cmBC = 7 \text{ cm}.
  2. Use a compass to draw an arc of 6 cm6 \text{ cm} from BB and an arc of 5 cm5 \text{ cm} from CC. The intersection point is AA. Join ABAB and ACAC.
  3. Construct the angle bisector of ∠B\angle B by drawing an arc and then two intersecting arcs from the points where the first arc cuts ABAB and BCBC.
  4. Similarly, construct the angle bisector of ∠C\angle C.
  5. The point where these two bisectors intersect is the incenter II.
  6. From II, draw a perpendicular to the side BCBC. Let the foot of the perpendicular be DD.
  7. With II as center and IDID as radius, draw the circle that touches all three sides.

Explanation:

The incenter is the equidistant point from all sides of the triangle. By bisecting the angles, we locate this point. The perpendicular distance to a side serves as the radius.

Problem 2:

Draw a circle of radius 3 cm3 \text{ cm}. From a point PP at a distance of 8 cm8 \text{ cm} from the center OO, construct two tangents to the circle. Measure the length of the tangents.

Solution:

  1. Draw a circle with center OO and radius 3 cm3 \text{ cm}.
  2. Mark a point PP such that OP=8 cmOP = 8 \text{ cm}.
  3. Construct the perpendicular bisector of OPOP: Draw arcs from OO and PP with radius greater than 4 cm4 \text{ cm} to find midpoint MM.
  4. With MM as center and MOMO (or MPMP) as radius, draw a dotted circle.
  5. Let the dotted circle intersect the original circle at points T1T_1 and T2T_2.
  6. Join PT1PT_1 and PT2PT_2. These are the required tangents.
  7. Calculation: PT1=OP2−OT12=82−32=64−9=55≈7.41 cmPT_1 = \sqrt{OP^2 - OT_1^2} = \sqrt{8^2 - 3^2} = \sqrt{64 - 9} = \sqrt{55} \approx 7.41 \text{ cm}.

Explanation:

This construction utilizes the property that the angle in a semi-circle is 90∘90^{\circ}. The dotted circle ensures that ∠OT1P\angle OT_1P is a right angle, making PT1PT_1 a tangent.

Problem 3:

Construct a triangle PQRPQR where PQ=5PQ = 5 cm, QR=6QR = 6 cm, and RP=7RP = 7 cm. Construct the circumcircle of triangle PQRPQR.

Circumcircle of triangle PQR with side lengths 5, 6, and 7 cm.

Solution:

  1. Draw a line QR=6QR = 6 cm.
  2. With QQ as center and radius 55 cm, draw an arc. With RR as center and radius 77 cm, draw another arc to intersect at PP.
  3. Join PQPQ and PRPR to form △PQR\triangle PQR.
  4. Draw the perpendicular bisectors of PQPQ and QRQR.
  5. Let the intersection of these bisectors be OO (the circumcenter).
  6. With OO as center and OPOP as radius, draw the circle passing through P,Q,P, Q, and RR.

Explanation:

The circumcenter is the point where the perpendicular bisectors of the sides meet. This point is equidistant from all vertices, making OP=OQ=OROP = OQ = OR.

Problem 4:

Construct a circle of radius 3.53.5 cm. Take a point MM on the circle and construct a tangent to the circle at point MM without using the center.

Construction of a tangent at point M using the Alternate Segment Theorem.

Solution:

  1. Draw a circle of radius 3.53.5 cm and mark point MM on the circumference.
  2. Draw any chord MNMN and take a third point LL on the major arc MNMN.
  3. Join MLML and NLNL to form △MLN\triangle MLN in the alternate segment.
  4. At point MM, construct an angle equal to ∠MNL\angle MNL (angles in alternate segments).
  5. The line forming this angle with chord MNMN is the required tangent.

Explanation:

According to the Alternate Segment Theorem, the angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment.