Tangents and Secants to Circles - 1 Mark Solutions
Important Geometric Principles:
• Tangent Length Formula: PT = √(OP² - OT²) where O is center, P is external point, T is tangent point
• Area of Sector: (θ/360°) × πr²
• Tangent is perpendicular to radius at point of contact
• Only one tangent can be drawn from a point on the circle
• Two tangents can be drawn from an external point to a circle
• Area of Sector: (θ/360°) × πr²
• Tangent is perpendicular to radius at point of contact
• Only one tangent can be drawn from a point on the circle
• Two tangents can be drawn from an external point to a circle
1
How many tangents can be drawn to a circle from a point on the same circle? Justify your answer.
(M'15)
Step 1: Understanding the concept
A tangent to a circle is a line that touches the circle at exactly one point.
Step 2: Property of tangents
From any point on the circle, there is exactly one line that is perpendicular to the radius at that point, and this line is the tangent.
Diagram:
Step 3: Conclusion
Since there is exactly one line perpendicular to the radius at any point on the circle, only one tangent can be drawn from a point on the circle.
∴ Only one tangent can be drawn to a circle from a point on the same circle.
2
Find the length of the tangent from a point, which is 9.1cm away from the centre of the circle, whose radius is 8.4cm.
(J'15)
Step 1: Visualize the right triangle
The tangent, radius, and line from center to external point form a right triangle with:
- Hypotenuse (OP) = 9.1 cm
- Radius (OT) = 8.4 cm
- Tangent (PT) = ?
- Hypotenuse (OP) = 9.1 cm
- Radius (OT) = 8.4 cm
- Tangent (PT) = ?
Step 2: Apply Pythagorean Theorem
(PT)² + (OT)² = (OP)²
(PT)² + (8.4)² = (9.1)²
(PT)² + 70.56 = 82.81
(PT)² + (8.4)² = (9.1)²
(PT)² + 70.56 = 82.81
Step 3: Solve for Tangent Length
(PT)² = 82.81 - 70.56 = 12.25
PT = √12.25 = 3.5 cm
PT = √12.25 = 3.5 cm
Interactive Calculator
∴ The length of the tangent is 3.5 cm.
3
The length of the tangent from an external point 'P' to a circle with center 'O' is always less than 'OP'. Is this statement true? Give reasons.
(J'16)
Step 1: Understanding the geometric relationship
In the right triangle OPT (where T is the point of tangency):
OT ⟂ PT (radius is perpendicular to tangent at point of contact)
OT ⟂ PT (radius is perpendicular to tangent at point of contact)
Step 2: Apply Pythagorean Theorem
(OP)² = (OT)² + (PT)²
Since (OT)² > 0, we have:
(OP)² > (PT)²
Therefore, OP > PT
Since (OT)² > 0, we have:
(OP)² > (PT)²
Therefore, OP > PT
Diagram:
∴ Yes, the statement is true. In the right triangle OPT, OP is the hypotenuse and PT is one of the legs, so OP > PT.
4
The length of the minute hand of a clock is 3.5 cm. Find the area swept by minute hand in 30 minutes. (use π = 22/7)
(M'17)
Step 1: Determine the angle swept
In 60 minutes, the minute hand completes a full circle (360°)
In 30 minutes, it sweeps 30/60 × 360° = 180° (a semicircle)
In 30 minutes, it sweeps 30/60 × 360° = 180° (a semicircle)
Step 2: Calculate area of semicircle
Area of full circle = πr² = (22/7) × (3.5)²
Area of semicircle = (1/2) × πr² = (1/2) × (22/7) × (3.5)²
Area of semicircle = (1/2) × πr² = (1/2) × (22/7) × (3.5)²
Step 3: Perform calculation
(1/2) × (22/7) × 12.25 = (1/2) × 22 × 1.75 = 11 × 1.75 = 19.25 cm²
Interactive Calculator
∴ The area swept by the minute hand in 30 minutes is 19.25 cm².
5
The length of the tangent to a circle from a point 17 cm from its Centre is 18 cm. Find the radius of the circle.
(M'18)
Step 1: Apply Pythagorean Theorem
In right triangle OPT:
(OP)² = (OT)² + (PT)²
(17)² = (r)² + (18)²
289 = r² + 324
(OP)² = (OT)² + (PT)²
(17)² = (r)² + (18)²
289 = r² + 324
Step 2: Solve for radius
r² = 324 - 289 = 35
r = √35 ≈ 5.92 cm
r = √35 ≈ 5.92 cm
Interactive Calculator
∴ The radius of the circle is √35 cm ≈ 5.92 cm.
6
Find the length of the tangent to circle from a point 13 cm away from the centre of the circle of radius 5 cm.
(J'18)
Step 1: Apply Pythagorean Theorem
In right triangle OPT:
(OP)² = (OT)² + (PT)²
(13)² = (5)² + (PT)²
169 = 25 + (PT)²
(OP)² = (OT)² + (PT)²
(13)² = (5)² + (PT)²
169 = 25 + (PT)²
Step 2: Solve for tangent length
(PT)² = 169 - 25 = 144
PT = √144 = 12 cm
PT = √144 = 12 cm
Interactive Calculator
∴ The length of the tangent is 12 cm.
7
A point P is 25 cm from the centre O of the circle. The length of the tangent drawn from P to the circle is 24 cm. Find the radius of the circle.
(M'19)
Step 1: Apply Pythagorean Theorem
In right triangle OPT:
(OP)² = (OT)² + (PT)²
(25)² = (r)² + (24)²
625 = r² + 576
(OP)² = (OT)² + (PT)²
(25)² = (r)² + (24)²
625 = r² + 576
Step 2: Solve for radius
r² = 625 - 576 = 49
r = √49 = 7 cm
r = √49 = 7 cm
Interactive Calculator
∴ The radius of the circle is 7 cm.
Similar Triangles Constructions -4 Marks Solutions
Important Geometric Principles:
• Tangent Length Formula: PT = √(OP² - OT²) where O is center, P is external point, T is tangent point
• Construction Steps: 1) Draw circle, 2) Mark external point, 3) Join to center, 4) Find midpoint, 5) Draw semicircle, 6) Mark intersection points, 7) Draw tangents
• For concentric circles: Chord of larger circle that touches smaller circle has length = 2√(R² - r²)
• Angle between tangents: If angle between tangents is θ, then angle between radii is 180° - θ
• Construction Steps: 1) Draw circle, 2) Mark external point, 3) Join to center, 4) Find midpoint, 5) Draw semicircle, 6) Mark intersection points, 7) Draw tangents
• For concentric circles: Chord of larger circle that touches smaller circle has length = 2√(R² - r²)
• Angle between tangents: If angle between tangents is θ, then angle between radii is 180° - θ
1
Draw a circle with radius 3cm and construct a pair of tangents from a point 8cm away from the centre.
(M'15)
Construction Steps:
- Draw a circle with center O and radius 3 cm.
- Mark a point P such that OP = 8 cm.
- Join O and P.
- Find the midpoint M of OP using perpendicular bisector construction.
- With M as center and MO as radius, draw a semicircle intersecting the circle at points T and T'.
- Join PT and PT'. These are the required tangents.
Step 2: Calculate tangent length
Using Pythagorean theorem:
Tangent length = √(OP² - r²) = √(8² - 3²) = √(64 - 9) = √55 ≈ 7.42 cm
Tangent length = √(OP² - r²) = √(8² - 3²) = √(64 - 9) = √55 ≈ 7.42 cm
Calculate Tangent Length
∴ The tangents PT and PT' have been constructed. Their length is √55 ≈ 7.42 cm.
2
Draw a circle of radius 5cm. from a point 8cm away from its centre, construct a pair of tangents to the circle. Find the lengths of tangents.
(M'16)
Construction Steps:
- Draw a circle with center O and radius 5 cm.
- Mark a point P such that OP = 8 cm.
- Join O and P.
- Find the midpoint M of OP using perpendicular bisector construction.
- With M as center and MO as radius, draw a semicircle intersecting the circle at points T and T'.
- Join PT and PT'. These are the required tangents.
Step 2: Calculate tangent length
Using Pythagorean theorem:
Tangent length = √(OP² - r²) = √(8² - 5²) = √(64 - 25) = √39 ≈ 6.24 cm
Tangent length = √(OP² - r²) = √(8² - 5²) = √(64 - 25) = √39 ≈ 6.24 cm
∴ The length of each tangent is √39 ≈ 6.24 cm.
3
Two concentric circles of radii 10cm and 6cm are drawn. Find the length of the chord of the larger circle which touches the smaller circle.
(J'16)
Step 1: Understand the geometry
Let the common center be O.
Let the chord of the larger circle be AB which touches the smaller circle at M.
Then OM ⟂ AB (radius is perpendicular to tangent at point of contact)
Let the chord of the larger circle be AB which touches the smaller circle at M.
Then OM ⟂ AB (radius is perpendicular to tangent at point of contact)
Step 2: Apply Pythagorean theorem
In right triangle OMA:
OA² = OM² + AM² (OA is radius of larger circle, OM is radius of smaller circle)
10² = 6² + AM²
100 = 36 + AM²
AM² = 64
AM = 8 cm
OA² = OM² + AM² (OA is radius of larger circle, OM is radius of smaller circle)
10² = 6² + AM²
100 = 36 + AM²
AM² = 64
AM = 8 cm
Step 3: Find chord length
Since M is the midpoint of chord AB (perpendicular from center bisects the chord):
AB = 2 × AM = 2 × 8 = 16 cm
AB = 2 × AM = 2 × 8 = 16 cm
Calculate Chord Length
∴ The length of the chord is 16 cm.
4
Draw a circle of diameter 6 cm from a point 5 cm away from its centre. Construct the pair of tangents to the circle and measure their length.
(M'17)
Construction Steps:
- Diameter = 6 cm, so radius = 3 cm. Draw a circle with center O and radius 3 cm.
- Mark a point P such that OP = 5 cm.
- Join O and P.
- Find the midpoint M of OP using perpendicular bisector construction.
- With M as center and MO as radius, draw a semicircle intersecting the circle at points T and T'.
- Join PT and PT'. These are the required tangents.
Step 2: Calculate tangent length
Using Pythagorean theorem:
Tangent length = √(OP² - r²) = √(5² - 3²) = √(25 - 9) = √16 = 4 cm
Tangent length = √(OP² - r²) = √(5² - 3²) = √(25 - 9) = √16 = 4 cm
∴ The length of each tangent is 4 cm.
5
Draw two concentric circles of radii 1.5 cm and 4 cm. From a point 10 cm from its centre, construct the pair of tangent to the circle.
(J'17)
Construction Steps:
- Draw two circles with common center O and radii 1.5 cm and 4 cm.
- Mark a point P such that OP = 10 cm.
- Join O and P.
- Find the midpoint M of OP using perpendicular bisector construction.
- With M as center and MO as radius, draw a semicircle intersecting the larger circle at points T and T'.
- Join PT and PT'. These are the required tangents to the larger circle.
Step 2: Calculate tangent length
Using Pythagorean theorem:
Tangent length = √(OP² - r²) = √(10² - 4²) = √(100 - 16) = √84 ≈ 9.17 cm
Tangent length = √(OP² - r²) = √(10² - 4²) = √(100 - 16) = √84 ≈ 9.17 cm
∴ The tangents to the larger circle have been constructed. Their length is √84 ≈ 9.17 cm.
6
Draw a circle of radius 6 cm and construct two tangents to the circle so that angle between the tangents is 60°.
(J'19)
Construction Steps:
- Draw a circle with center O and radius 6 cm.
- Draw a radius OA.
- At O, construct ∠AOB = 120° (since angle between tangents is 60°, the angle between radii is 180° - 60° = 120°).
- At A and B, draw lines perpendicular to OA and OB respectively. These lines are the tangents and they will intersect at point P.
- The angle between the tangents PA and PB will be 60°.
Step 2: Calculate distance OP
In right triangle OAP:
OP = OA / cos(∠AOP) = 6 / cos(60°) = 6 / 0.5 = 12 cm
So the external point P is 12 cm from the center O.
OP = OA / cos(∠AOP) = 6 / cos(60°) = 6 / 0.5 = 12 cm
So the external point P is 12 cm from the center O.
∴ The tangents PA and PB have been constructed with an angle of 60° between them. The external point P is 12 cm from the center O.
7
Draw a circle of radius 4 cm. From a point 9 cm away from it's centre, construct a pair of tangents to the circle.
(May 2022)
Construction Steps:
- Draw a circle with center O and radius 4 cm.
- Mark a point P such that OP = 9 cm.
- Join O and P.
- Find the midpoint M of OP using perpendicular bisector construction.
- With M as center and MO as radius, draw a semicircle intersecting the circle at points T and T'.
- Join PT and PT'. These are the required tangents.
Step 2: Calculate tangent length
Using Pythagorean theorem:
Tangent length = √(OP² - r²) = √(9² - 4²) = √(81 - 16) = √65 ≈ 8.06 cm
Tangent length = √(OP² - r²) = √(9² - 4²) = √(81 - 16) = √65 ≈ 8.06 cm
∴ The tangents have been constructed. Their length is √65 ≈ 8.06 cm.
8
Construct a circle of radius 5 cm. Then construct a pair of tangents to the circle such that the angle between them is 60°.
(Jun'23)
Construction Steps:
- Draw a circle with center O and radius 5 cm.
- Draw a radius OA.
- At O, construct ∠AOB = 120° (since angle between tangents is 60°, the angle between radii is 180° - 60° = 120°).
- At A and B, draw lines perpendicular to OA and OB respectively. These lines are the tangents and they will intersect at point P.
- The angle between the tangents PA and PB will be 60°.
Step 2: Calculate distance OP
In right triangle OAP:
OP = OA / cos(∠AOP) = 5 / cos(60°) = 5 / 0.5 = 10 cm
So the external point P is 10 cm from the center O.
OP = OA / cos(∠AOP) = 5 / cos(60°) = 5 / 0.5 = 10 cm
So the external point P is 10 cm from the center O.
∴ The tangents PA and PB have been constructed with an angle of 60° between them. The external point P is 10 cm from the center O.
