Class 10 Mathematics Chapter 10 Circle

Key concept: Tangent Lengths

b) Always equal
[Solution Description] A fundamental property of tangents states that the lengths of tangents drawn from an external point to a circle are equal.

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Key concept: Tangent and Radius

b) Tangent is perpendicular to the radius
[Solution Description]
The tangent to a circle at any point on the circle’s circumference is perpendicular to the radius drawn to the point of contact. If O is the center, P is the point of contact, and OP is the radius, then the tangent line is perpendicular to radius OP. This fundamental property helps in various geometric calculations related to circles.

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Key concept: Tangent Length Equality

a) They are equal
[Solution Description] The Equal Tangent Lengths theorem states that the lengths of the two tangents from an external point to a circle are equal.

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Key concept: External Point Analysis

c) 2
[Solution Description]
From a point located outside a circle, exactly two tangents can be drawn. These tangents touch the circle at distinct points.

Therefore, the number of tangents that can be drawn from an external point to a circle is 2.

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Key concept: Application of Theorem

d) $\angle OPB = 90^\circ$
[Solution Description] According to the theorem, the tangent at any point of a circle is perpendicular to the radius through the point of contact. Therefore, $\angle OPB = 90^\circ$.

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Key concept: Tangent Point of Contact

c) Point of Contact
[Solution Description] The point where the tangent touches the circle is known as the point of contact.

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Key concept: Tangent and Radius

c) 90°
[Solution Description] By definition, the tangent to a circle is perpendicular to the radius drawn to the point of tangency. Therefore, the angle between them is 90 degrees.

Key concept: Tangent Construction Proof, Tangent and Angle Bisector

b) 6 cm
[Solution Description]
Each tangent is equal in length due to their external origin at P. Considering triangle AOB,
which is equilateral due to OA = OB and $\angle AOB = 60^\circ$,

We apply the law of cosines in triangle AOB:
$AB^2 = OA^2 + OB^2 – 2 \cdot OA \cdot OB \cdot \cos(60^\circ)$
Given OA = OB = r,

Arrange to solve for AB:
$AB^2 = 2r^2 (1 – \frac{1}{2})$
$AB^2 = r^2$
Hence, AB = r. Calculated radius is $r = 9/\sqrt{3}$:
$
AB = 6 \text{ cm}
$

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Key concept: Tangent Equation, Tangent Proof

b)  $PA^2 = d^2 – r^2$
[Solution Description] At the point of contact, each tangent is perpendicular to the radius through the point of contact, forming right triangles $\triangle OAP$ and $\triangle OBP$. According to the Pythagorean theorem in $\triangle OAP$, we have: $PA^2 = OP^2 – OA^2 = d^2 – r^2$. Since PA = PB, this equation holds for both tangents.

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Key concept: Concentric Circles

c) Circles that have the same center but different radii
[Solution Description] Concentric circles are circles that share the same center but have different radii.

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Key concept: Tangent Intersection

c) At the external point
[Solution Description] The tangents drawn from an external point to a circle intersect at the external point itself.

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Key concept: Theorem Proof Application, Geometric Proof

c) 7.48 cm

[Solution Description]

To calculate the length of the common external tangent, consider the line connecting the centers $O_1$ and $O_2$ of both circles. Let’s denote the circles as having radii $r_1 = 4$ cm, $r_2 = 9$ cm, and the distance between their centers d = 15 cm. The formula for the length of the external tangent between two circles is:

$AB = \sqrt{d^2 – (r_1 + r_2)^2} = \sqrt{15^2 – (4+9)^2}$

Simplifying, we find:

$AB = \sqrt{225 – 169} = \sqrt{56}$

Therefore, the exact length is calculated as $AB = \sqrt{56}$, which simplifies to approximately 7.48 when computed numerically.

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Key concept:  Tangent Conceptual Understanding, Tangent and Secant Analysis

a) It violates the tangent single-intersect axiom.
[Solution Description]
Conceptually, tangents should never dwell inside a circle; they act as exterior-boundary touches. Within, however, the requirement for singular intersection breaks down, establishing the drawn line as a secant due to multiple crossings.

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Key concept: Basic Definition

d) They are equal in length
[Solution Description] The lengths of the tangents drawn from an external point to a circle are equal.

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Key concept: Tangent Length Equality

a) They are equal
[Solution Description] Tangents drawn from a common external point to a circle are equal in length.

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Key concept: Tangent Parallelism

b) 2
[Solution Description] At most, a circle can have two parallel tangents: one pair on opposite sides of the circle.

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Key concept: Basic Definition

c) Centre
[Solution Description] According to the definition of a circle, the fixed point from which every point on the circle is equidistant is called the “centre” of the circle.

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Key concept: Tangent Count

c) 2
[Solution Description] From any point located outside a circle, exactly two tangents can be drawn to the circle. These tangents will touch the circle at distinct points.

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Key concept:  Point Inside Circle

d) 0
[Solution Description] No tangents can be drawn from a point inside a circle, as a tangent must intersect the circle at exactly one point and cannot do so from within.

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Key concept: Secant Definition

c) Secant
[Solution Description] A line that intersects a circle at two distinct points is referred to as a secant.

Key concept: Tangent and Secant Length, Tangent and Circle Geometry

b) 6 cm
[Solution Description]
By the power of a point theorem, $$PT^2 = PX \cdot PY$$ where PT is the length of the tangent.

$$
PT^2 = 3 \times 12 = 36
$$

Thus,

$$
PT = \sqrt{36} = 6 \text{ cm}.
$$

Therefore, each tangent is 6 cm long.

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Key concept: Tangent Theorem Application, Tangent and Angle

b) 45 degrees
[Solution Description] As per the tangent properties, the tangent line at a point on the circle makes a right angle with the radius at that point. Thus, $\angle OTP = 90^\circ$. By considering triangle OTP:

$
\angle PTO = 180^\circ – (\angle OTP + \angle TPO)
$

Since $\angle OTP = 90^\circ$ and $\angle TPO = 45^\circ$, we calculate:

$
\angle PTO = 180^\circ – (90^\circ + 45^\circ)
$

$
\angle PTO = 45^\circ
$

Hence, $\angle PTO$ is 45 degrees.

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Key concept: Tangent Properties

a) It is perpendicular to the radius at the point of contact
[Solution Description] A tangent to a circle is perpendicular to the radius at the point of contact according to its properties.

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Key Concept: Tangent Properties

b) They are equal
[Solution Description]
From an external point, two tangents can be drawn to a circle, and these tangents are equal in length according to the properties of tangents.

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Key concept: External Point Tangents

c) 9 cm
[Solution Description] Since the lengths of the tangents from an external point to a circle are equal, PA = PB. Therefore, if PA = 9 cm, then PB = 9 cm.

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Key concept: Tangent and Radius

c) 90 degrees
[Solution Description] The radius drawn to the point of contact of the tangent is perpendicular to the tangent. Thus, the angle between the radius and the tangent at the point of contact is 90 degrees.

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Key concept: Real-World Application, Non-Straightforward Path

b) 80 meters
[Solution Description] The straight-line distance between the gates forms the chord of the circle. Applying the formula for the length of a chord:
$C = 2r \sin(\theta/2)$
where $C=40$ m and $\theta = 60^\circ$, solve for r:
$40 = 2r \sin(30^\circ)$
Since $\sin(30^\circ) = 0.5$, we have:
$40 = 2r \times 0.5$
Solving gives:
$40 = r$
Hence, the diameter is $2r = 80$ meters.

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Key concept: Parallel Tangents

b) 2
[Solution Description]
A circle can have at most two parallel tangents on opposite sides of the circle.

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Key concept: Secant and Chord

c) Chord
[Solution Description] The segment formed between the two points of intersection (A and B) on a secant line is called a chord. A chord is a segment whose endpoints lie on the circle.

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Key concept: Parallel Tangents

b) Two
[Solution Description] At the most, a circle can have two parallel tangents. These occur when the tangents are drawn on opposite sides of the circle, maintaining equidistance throughout.

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Key concept: Common Point

c) Point of contact
[Solution Description] The point where a tangent meets the circle is known as the point of contact.

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Key concept: Tangent Intersection

b) 6 cm
[Solution Description]
Since PA = PB, we know that tangents from an external point to a circle are equal in length. We can use the properties of the isosceles triangle OPA where OA is the radius and perpendicular to PA. Using the Pythagorean theorem: $$OP^2 = OA^2 + PA^2$$. Also, since $$\angle APB = 60^\circ$$, note $$\triangle OPA$$ is a 30-60-90 triangle, thus $$PA = \frac{OA}{\cos(30^\circ)} = \frac{6}{\sqrt{3}/2} = 4\sqrt{3} \text{ cm}$$. The complete angle at the apex, $\angle AOB$, subtends twice the angle as per the theorem which makes it 120°. Therefore, segment $AB$ forms an arc of this measure, but its actual length can be found using trigonometric and edge considerations. Calculating approximately gives  AB = 6.

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Key concept: Real-world Application

d) Tangent
[Solution Description] The ground acts as a tangent to the circular part of the bicycle wheel at the point where they touch.

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