Pythagorean Theorem: a² + b² = c² (for right triangles)
Tangent Properties: Tangent is perpendicular to radius at point of contact
Inradius Formula: For right triangle, r = (a + b - c)/2
1 Tangent Length and Radius
The length of the tangent to a circle from a point P, which is 25 cm away from the centre is 24 cm. What is the radius of the circle?
1 We know that the tangent to a circle is perpendicular to the radius at the point of contact.
2 This forms a right triangle with the radius (r), tangent (24 cm), and distance from point to center (25 cm).
3 Using Pythagoras theorem: r² + 24² = 25²
4 Calculate: r² = 625 - 576 = 49
5 Therefore, r = √49 = 7 cm
Answer: The radius of the circle is 7 cm
2 Inscribed Circle in Right Triangle
△LMN is a right angled triangle with ∠L = 90°. A circle is inscribed in it. The lengths of the sides containing the right angle are 6 cm and 8 cm. Find the radius of the circle.
1 First, find the hypotenuse using Pythagoras theorem: √(6² + 8²) = 10 cm
2 The formula for radius of incircle in right triangle is r = (a + b - c)/2, where c is hypotenuse
3 Plug in values: r = (6 + 8 - 10)/2 = 4/2 = 2 cm
Answer: The radius of the inscribed circle is 2 cm
3 Incircle of Triangle ABC
A circle is inscribed in △ABC having sides 8 cm, 10 cm and 12 cm. Find AD, BE and CF.
1 Let AD = AF = x, BD = BE = y, CE = CF = z
2 We know: x + y = 8, y + z = 10, z + x = 12
3 Add all three equations: 2(x + y + z) = 30 ⇒ x + y + z = 15
4 Subtract each original equation from this:
z = 15 - 8 = 7 cm
x = 15 - 10 = 5 cm
y = 15 - 12 = 3 cm
5 Therefore:
AD = x = 5 cm
BE = y = 3 cm
CF = z = 7 cm
Answer: AD = 5 cm, BE = 3 cm, CF = 7 cm
4 Tangent and Angle
PQ is a tangent drawn from a point P to a circle with centre O and QOR is a diameter of the circle such that ∠POR = 120°. Find ∠OPQ.
1 Since QOR is diameter and ∠POR = 120°, then ∠POQ = 60° (angles on straight line)
2 OP = OQ (both radii), so △OPQ is isosceles
3 Therefore, ∠OPQ = ∠OQP = (180° - 60°)/2 = 60°
4 But PQ is tangent, so ∠OQP = 90° (radius perpendicular to tangent)
5 Therefore, ∠OPQ = 180° - 90° - 60° = 30°
Answer: ∠OPQ = 30°
5 Tangent and Chord Angle
A tangent ST to a circle touches it at B. AB is a chord such that ∠ABT = 65°. Find ∠AOB, where "O" is the centre of the circle.
1 ∠OBT = 90° (radius perpendicular to tangent)
2 Therefore, ∠OBA = 90° - 65° = 25°
3 OA = OB (both radii), so △OAB is isosceles
4 Therefore, ∠OAB = ∠OBA = 25°
5 Thus, ∠AOB = 180° - 2×25° = 130°
Answer: ∠AOB = 130°
6 Tangent Length Calculation
In figure, O is the centre of the circle with radius 5 cm. T is a point such that OT = 13 cm and OT intersects the circle at E, if AB is the tangent to the circle at E, find the length of AB.
1 First find length of tangent from T to circle: √(13² - 5²) = √(169 - 25) = √144 = 12 cm
2 AB is tangent at E, so it's perpendicular to OE (radius)
3 Using similar triangles or power of point, AB = 2 × (5 × 12)/13 ≈ 9.23 cm
Answer: AB ≈ 9.23 cm
7 Concentric Circles
In two concentric circles, a chord of length 16 cm of larger circle becomes a tangent to the smaller circle whose radius is 6 cm. Find the radius of the larger circle.
1 The chord of the larger circle is tangent to the smaller circle, so it's perpendicular to the radius of the smaller circle.
2 The chord is bisected by this radius, forming two right triangles with legs 6 cm (radius of small circle) and 8 cm (half the chord).
Two circles with centres O and O' of radii 3 cm and 4 cm, respectively intersect at two points P and Q, such that OP and O'P are tangents to the two circles. Find the length of the common chord PQ.
1 Since OP is tangent to O' circle, O'P ⊥ OP
2 Similarly, O'P is tangent to O circle, so OP ⊥ O'P
3 Therefore, OPO' is right triangle with right angle at P
4 Let OO' = d, then d² = 3² + 4² = 25 ⇒ d = 5 cm
5 Common chord PQ is perpendicular to OO' and bisected by it
6 Using right triangle properties, PQ = 2 × (3×4)/5 = 4.8 cm
Answer: The length of common chord PQ is 4.8 cm
9 Angle Bisectors Concurrency
Show that the angle bisectors of a triangle are concurrent.
1 Consider △ABC with angle bisectors of ∠A and ∠B intersecting at I
2 Since I lies on bisector of ∠A, it's equidistant from AB and AC
3 Since I lies on bisector of ∠B, it's equidistant from AB and BC
4 Therefore, I is equidistant from AC and BC, so must lie on bisector of ∠C
5 Thus all three angle bisectors meet at I (incenter)
Answer: The angle bisectors meet at the incenter (concurrent)
10 Triangular Window Problem
An artist has created a triangular stained glass window and has one strip of small length left before completing the window. She needs to figure out the length of left out portion based on the lengths of the other sides as shown in the figure.
1 This problem requires using the triangle inequality theorem
2 Sum of any two sides must be greater than third side
3 If two sides are a and b, then third side x must satisfy:
|a - b| < x < a + b
4 For example, if sides are 5 cm and 7 cm, missing side must be between 2 cm and 12 cm
Answer: The missing length must satisfy triangle inequality conditions
11 Drawing Tangent at Point
Draw a tangent at any point R on the circle of radius 3.4 cm and centre at P.
1 Draw circle with center P and radius 3.4 cm
2 Mark any point R on the circumference
3 Join PR (radius)
4 At point R, draw line perpendicular to PR
5 This perpendicular line is the required tangent
Answer: The tangent is constructed perpendicular to radius PR at point R
12 Tangent Using Alternate Segment
Draw a circle of radius 4.5 cm. Take a point on the circle. Draw the tangent at that point using the alternate segment theorem.
1 Draw circle with radius 4.5 cm and center O
2 Mark point P on circumference and draw chord PQ
3 Draw angle in alternate segment equal to angle between chord and tangent
4 The line making this angle with PQ is the required tangent
Answer: Tangent is constructed using alternate segment theorem
13 Tangents from External Point
Draw the two tangents from a point which is 10 cm away from the centre of a circle of radius 5 cm. Also, measure the lengths of the tangents.
1 Draw circle with center O and radius 5 cm
2 Mark point P at 10 cm from O
3 Find midpoint M of OP (since OP = 10 cm, OM = 5 cm)
4 With M as center and OM as radius, draw semicircle intersecting original circle at T and T'
5 Join PT and PT', which are the required tangents
6 Length of tangents = √(10² - 5²) = √75 ≈ 8.66 cm
Answer: Two tangents constructed, each ≈ 8.66 cm long
14 Tangents from Distant Point
Take a point which is 11 cm away from the centre of a circle of radius 4 cm and draw the two tangents to the circle from that point.
1 Draw circle with center O and radius 4 cm
2 Mark point P at 11 cm from O
3 Find midpoint M of OP (since OP = 11 cm, OM = 5.5 cm)
4 With M as center and OM as radius, draw semicircle intersecting original circle at T and T'
5 Join PT and PT', which are the required tangents
6 Length of tangents = √(11² - 4²) = √105 ≈ 10.25 cm
Answer: Two tangents constructed, each ≈ 10.25 cm long
15 Tangents to Diameter Circle
Draw the two tangents from a point which is 5 cm away from the centre of a circle of diameter 6 cm. Also, measure the lengths of the tangents.
1 Draw circle with center O and radius 3 cm (since diameter = 6 cm)
2 Mark point P at 5 cm from O
3 Find midpoint M of OP (since OP = 5 cm, OM = 2.5 cm)
4 With M as center and OM as radius, draw semicircle intersecting original circle at T and T'
5 Join PT and PT', which are the required tangents
6 Length of tangents = √(5² - 3²) = √16 = 4 cm
Answer: Two tangents constructed, each 4 cm long
16 Drawing a Tangent
Draw a tangent to the circle from the point P having radius 3.6 cm, and centre at O. Point P is at a distance 7.2 cm from the centre.
1 Draw circle with center O and radius 3.6 cm
2 Mark point P at 7.2 cm from O (exactly twice the radius)
3 Find midpoint M of OP (since OP = 7.2 cm, OM = 3.6 cm)
4 With M as center and OM as radius, draw semicircle intersecting original circle at T