NCERT Class 10 Maths Solutions: Circles

Question:

Tangents from an external point to a circle are equally inclined to the line joining the point to the centre. Also, the radius to the point of tangency is perpendicular to the tangent.


Given that tangents PA and PB from point P to a circle with centre O are inclined to each other at an angle of 80°. This means ∠APB = 80°.
Consider the quadrilateral PAOB. The sum of angles in a quadrilateral is 360°.
We know that the radius to the point of tangency is perpendicular to the tangent. Therefore, ∠OAP = 90° and ∠OBP = 90°.
In quadrilateral PAOB, ∠APB + ∠OAP + ∠AOB + ∠OBP = 360°.
Substituting the known values: 80° + 90° + ∠AOB + 90° = 360°.
180° + ∠AOB + 80° = 360°.
∠AOB + 260° = 360°.
∠AOB = 360° – 260° = 100°.

Now, consider the triangles ΔPOA and ΔPOB.
OA = OB (radii of the same circle)
OP = OP (common side)
PA = PB (tangents from an external point to a circle are equal in length)
Therefore, ΔPOA ≅ ΔPOB by SSS congruence rule.
Since the triangles are congruent, their corresponding angles are equal.
So, ∠POA = ∠POB.
Also, ∠APB is bisected by OP, so ∠APO = ∠BPO = 80°/2 = 40°.
In right-angled triangle ΔOAP, the sum of angles is 180°.
∠OAP + ∠APO + ∠POA = 180°.
90° + 40° + ∠POA = 180°.
130° + ∠POA = 180°.
∠POA = 180° – 130° = 50°.
Alternatively, since ∠AOB = 100° and ∠POA = ∠POB, we have ∠POA = ∠AOB / 2 = 100° / 2 = 50°.

The final answer is $\boxed{50}$.

Question:

The question involves a circle, a tangent, and a line segment from the center to the point where the tangent meets the line. The key concept here is the property of a tangent to a circle: the radius drawn to the point of contact is perpendicular to the tangent. This forms a right-angled triangle, allowing us to use the Pythagorean theorem.


We are given a circle with center O and radius 5 cm.
PQ is a tangent to the circle at point P. This means that the radius OP is perpendicular to the tangent PQ.
Therefore, triangle OPQ is a right-angled triangle with the right angle at P.
We are given the radius OP = 5 cm and the distance OQ = 12 cm.
In the right-angled triangle OPQ, OQ is the hypotenuse (since it is opposite the right angle at P).
According to the Pythagorean theorem:
$OP^2 + PQ^2 = OQ^2$
Substitute the given values:
$5^2 + PQ^2 = 12^2$
$25 + PQ^2 = 144$
Subtract 25 from both sides:
$PQ^2 = 144 – 25$
$PQ^2 = 119$
Take the square root of both sides to find the length of PQ:
$PQ = \sqrt{119}$ cm

Therefore, the length of PQ is $\sqrt{119}$ cm.

The final answer is $\boxed{\sqrt{119}}$.

Question:

The relationship between the radius of a circle, the length of a tangent from an external point, and the distance of that point from the center of the circle forms a right-angled triangle. The Pythagorean theorem is applied to this triangle.


Let O be the center of the circle, P be the point of tangency on the circle, and Q be the external point.
We are given:
Length of the tangent from Q to the circle (QP) = 24 cm
Distance of Q from the center (OQ) = 25 cm
We need to find the radius of the circle (OP).

The radius drawn to the point of tangency is perpendicular to the tangent. Therefore, triangle OPQ is a right-angled triangle, with the right angle at P.

According to the Pythagorean theorem:
(Hypotenuse)^2 = (Perpendicular)^2 + (Base)^2
In triangle OPQ, OQ is the hypotenuse, OP is the radius (one leg), and QP is the tangent length (the other leg).

So, OQ^2 = OP^2 + QP^2
Substitute the given values:
25^2 = OP^2 + 24^2
625 = OP^2 + 576
OP^2 = 625 – 576
OP^2 = 49
OP = sqrt(49)
OP = 7 cm

Therefore, the radius of the circle is 7 cm.

Final Answer: The final answer is $\boxed{7}$

Question:

Tangents to a circle are perpendicular to the radius through the point of contact. The definition of a circle is the set of all points equidistant from a central point. The perpendicular bisector of a chord passes through the center of the circle.


Let the circle have center O and radius r. Let AB be the tangent to the circle at point P.
We need to prove that the perpendicular at P to AB passes through O.

Let’s assume that the perpendicular to AB at P does not pass through O.
Let this perpendicular line be L. So, L is perpendicular to AB at P.
Since we assumed L does not pass through O, O is not on L.

Draw a line segment OP. We know that OP is the radius of the circle.
By the theorem, the radius through the point of contact of a tangent is perpendicular to the tangent.
Therefore, OP is perpendicular to AB at P.

Now we have two lines, L and OP, both passing through P and both perpendicular to the same line AB.
In Euclidean geometry, through a given point (P) and a given line (AB), there can be only one unique perpendicular line.
This means that L and OP must be the same line.

Since OP is the line passing through P and perpendicular to AB, and we assumed L is also the line passing through P and perpendicular to AB, L must be the same line as OP.
The line OP passes through the center O by definition.
Therefore, the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Question:

Tangents to a circle from an external point, properties of quadrilaterals, angles in a circle, cyclic quadrilaterals.


Let O be the center of the circle and P be an external point. Let the tangents from P touch the circle at points A and B respectively. Join OA, OB, and AB. We need to prove that the angle between the tangents (angle APB) and the angle subtended by the line segments joining the points of contact to the centre (angle AOB) are supplementary, i.e., angle APB + angle AOB = 180 degrees.

We know that the radius drawn to the point of contact is perpendicular to the tangent. Therefore, angle OAP = 90 degrees and angle OBP = 90 degrees.

Consider the quadrilateral PAOB. The sum of the interior angles of a quadrilateral is 360 degrees.
So, in quadrilateral PAOB, we have:
angle APB + angle PAO + angle AOB + angle OBP = 360 degrees

Substitute the known values:
angle APB + 90 degrees + angle AOB + 90 degrees = 360 degrees

Simplify the equation:
angle APB + angle AOB + 180 degrees = 360 degrees

Subtract 180 degrees from both sides:
angle APB + angle AOB = 360 degrees – 180 degrees
angle APB + angle AOB = 180 degrees

This proves that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segments joining the points of contact to the centre.

Question:

Pythagorean Theorem: In a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. The relationship is given by $a^2 + b^2 = c^2$, where ‘c’ is the hypotenuse and ‘a’ and ‘b’ are the other two sides.

Radius and Tangent Property: A tangent to a circle is perpendicular to the radius drawn to the point of contact. This forms a right-angled triangle.


Let O be the center of the circle and A be the external point.
Let P be the point of contact of the tangent from A to the circle.
We are given that the distance of point A from the center O is 5 cm. So, OA = 5 cm.
We are given that the length of the tangent from point A to the circle is 4 cm. So, AP = 4 cm.
According to the property of tangents, the radius drawn to the point of contact is perpendicular to the tangent. Therefore, the angle OPA is a right angle (90 degrees).
This means that triangle OPA is a right-angled triangle, with OA as the hypotenuse.
We can apply the Pythagorean Theorem to triangle OPA:
$OP^2 + AP^2 = OA^2$
Let ‘r’ be the radius of the circle, which is the length of OP.
So, $r^2 + 4^2 = 5^2$
$r^2 + 16 = 25$
$r^2 = 25 – 16$
$r^2 = 9$
$r = \sqrt{9}$
$r = 3$ cm.
Therefore, the radius of the circle is 3 cm.

Question:

Pythagorean Theorem, properties of tangents to a circle, and geometric interpretation of concentric circles.


Let the two concentric circles have center O. Let the radius of the larger circle be R and the radius of the smaller circle be r.
Given R = 5 cm and r = 3 cm.
Let AB be the chord of the larger circle that touches the smaller circle at point P.
Since AB is a chord of the larger circle and also a tangent to the smaller circle at P, OP is perpendicular to AB. Thus, OP is the radius of the smaller circle and OP = r = 3 cm.
Also, OP bisects the chord AB because a perpendicular from the center to a chord bisects the chord. Therefore, AP = PB.
Consider the right-angled triangle OPA.
OA is the radius of the larger circle, so OA = R = 5 cm.
OP is the radius of the smaller circle, so OP = r = 3 cm.
Using the Pythagorean theorem in triangle OPA:
OA^2 = OP^2 + AP^2
5^2 = 3^2 + AP^2
25 = 9 + AP^2
AP^2 = 25 – 9
AP^2 = 16
AP = sqrt(16)
AP = 4 cm

Since AB = AP + PB and AP = PB, we have AB = 2 * AP.
AB = 2 * 4 cm
AB = 8 cm

Therefore, the length of the chord of the larger circle which touches the smaller circle is 8 cm.

Question:

A tangent to a circle is a line that touches the circle at exactly one point. Parallel lines are lines that never intersect, no matter how far they are extended. To find the maximum number of parallel tangents to a circle, we need to consider the points on the circle where a tangent line can be drawn and how these tangent lines relate to each other geometrically.


A circle is a symmetrical shape. Consider a point P on the circle. The tangent at point P is perpendicular to the radius OP. If we consider a point diametrically opposite to P, let’s call it Q, the tangent at Q will be perpendicular to the radius OQ. Since OP and OQ are along the same line (a diameter), and both tangents are perpendicular to this diameter, the tangents at P and Q will be parallel to each other. Any other tangent to the circle will not be parallel to the tangent at P or Q because the radius to the point of tangency will not be parallel to OP or OQ. Therefore, a circle can have at most two parallel tangents.

Question:

A circle is a set of all points in a plane that are at a fixed distance from a fixed point. A line is a straight one-dimensional figure that has no thickness and extends infinitely in both directions. When a line interacts with a circle, it can do so in a few ways. The question specifically asks about a line that cuts through the circle at two distinct locations. This geometric interaction has a specific name.


When a straight line passes through a circle and intersects its boundary at exactly two distinct points, it is defined as a secant. The line segment connecting these two points of intersection on the circle is called a chord. However, the question is about the line itself, not the segment within the circle. Therefore, the correct term for a line intersecting a circle in two points is a secant.

Question:

A tangent to a circle is a line that touches the circle at exactly one point. This point of contact is called the point of tangency.


A tangent to a circle is defined as a line that intersects the circle at precisely one point. Imagine a line “grazing” the edge of the circle without going inside it. That single point where the line meets the circle is the point of tangency. If a line intersects a circle at two points, it is called a secant. If it does not intersect the circle at all, it is simply a line outside the circle. Therefore, a tangent intersects a circle in exactly one point.

Answer: one

Question:

A tangent to a circle is a line that touches the circle at exactly one point. This point is called the point of contact.


Consider a circle. For any point on the circumference of the circle, we can draw a line that is perpendicular to the radius at that point. This line will touch the circle at only that single point and will not intersect the circle at any other point. Since there are infinitely many points on the circumference of a circle, we can draw a tangent at each of these points. Therefore, a circle can have infinitely many tangents.

Question:

The question requires understanding the definition of a tangent to a circle. A tangent is a line that touches the circle at exactly one point.


A tangent to a circle is a line that intersects the circle at precisely one point. This unique point where the tangent line touches the circle is called the point of contact or the point of tangency.

Therefore, the common point of a tangent to a circle and the circle is called the point of contact.