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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

Tangent to a Circle: A tangent is a straight line that touches the circumference of a circle at exactly one point, known as the point of contact. A key visual property is that the radius of the circle is always perpendicular to the tangent at the point of contact, forming a 9090^{\circ} angle.

Tangents from an External Point: From any point outside a circle, exactly two tangents can be drawn to the circle. These two tangents are equal in length (PA=PBPA = PB where PP is the external point). Visually, the line joining the center of the circle to the external point bisects the angle between the two tangents.

Circumscribed Circle (Circumcircle): This is a circle that passes through all three vertices of a triangle. The center of this circle, called the circumcenter, is found by constructing the perpendicular bisectors of at least two sides of the triangle. The distance from the circumcenter to any vertex is the circumradius (RR).

Inscribed Circle (Incircle): This is a circle that lies inside a triangle and touches all three of its sides. The center of this circle, called the incenter, is found by constructing the angle bisectors of at least two internal angles of the triangle. The perpendicular distance from the incenter to any side is the inradius (rr).

Construction of Tangent at a Point on the Circle: To construct a tangent at a point PP on the circumference, draw the radius OPOP. Then, using a compass, construct a line perpendicular to OPOP at point PP. This line is the required tangent.

Construction of Tangents from an External Point: To draw tangents from external point PP to a circle with center OO, first join OPOP. Construct the perpendicular bisector of OPOP to find its midpoint MM. Draw a second circle with center MM and radius MOMO. The points where this second circle intersects the original circle are the points of contact for the tangents from PP.

Regular Hexagon in a Circle: To inscribe a regular hexagon in a circle, the radius of the circle is used as the side length. By placing the compass pointer on the circumference and marking off arcs equal to the radius, exactly six points are created. Connecting these points forms a hexagon where each internal angle is 120120^{\circ}.

📐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=PAPBPT^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}.\n2. 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.\n3. 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.\n4. Similarly, construct the angle bisector of C\angle C.\n5. The point where these two bisectors intersect is the incenter II.\n6. From II, draw a perpendicular to the side BCBC. Let the foot of the perpendicular be DD.\n7. 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}.\n2. Mark a point PP such that OP=8 cmOP = 8 \text{ cm}.\n3. Construct the perpendicular bisector of OPOP: Draw arcs from OO and PP with radius greater than 4 cm4 \text{ cm} to find midpoint MM.\n4. With MM as center and MOMO (or MPMP) as radius, draw a dotted circle.\n5. Let the dotted circle intersect the original circle at points T1T_1 and T2T_2.\n6. Join PT1PT_1 and PT2PT_2. These are the required tangents.\n7. Calculation: PT1=OP2OT12=8232=649=557.41 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 9090^{\circ}. The dotted circle ensures that OT1P\angle OT_1P is a right angle, making PT1PT_1 a tangent.