Circles Class 10: NCERT Concepts and Theorems

Key Terms for Circles: A Quick Refresher

Circle

A collection of all points in a plane that are at a fixed distance (the radius) from a fixed point (the center).

Secant

A line that intersects a circle at two distinct points. It passes through the interior of the circle.

Tangent

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

Point of Contact

The single common point between a tangent and a circle.

Understanding Tangents to a Circle

Imagine a circle and a straight line on the same plane. There are only three possibilities for their interaction. First, the line might not touch the circle at all (a non-intersecting line). Second, the line might cut through the circle at two points; we call this line a secant. The third and most interesting case for us is when the line just grazes the circle, touching it at a single, unique point. This special line is called a tangent. Think of a bicycle wheel rolling on a straight path; the path acts as a tangent to the wheel at the point it touches the ground. An important property to remember is that for any given point on a circle, there can be only one tangent that passes through it. From a point outside the circle, you can draw exactly two tangents to the circle. From a point inside the circle, you cannot draw any tangents.

The Two Crucial Theorems on Tangents

Applying the Theorems: Solved Examples

• Example 1: Using Theorem 10.1 and Pythagoras Theorem A tangent PQ at a point P of a circle of radius 5 cm meets a line through the center O at a point Q so that OQ = 13 cm. Find the length of the tangent PQ. Step 1: Visualize and Draw Draw a circle with center O and radius OP = 5 cm. Draw a tangent PQ at point P. The line from O meets this tangent at Q, with OQ = 13 cm. Step 2: Apply Theorem 10.1 According to Theorem 10.1, the radius is perpendicular to the tangent at the point of contact. Therefore, OP ⊥ PQ. This means that △OPQ is a right-angled triangle, with the right angle at P. Step 3: Use Pythagoras Theorem In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Here, OQ is the hypotenuse. OQ² = OP² + PQ² 13² = 5² + PQ² 169 = 25 + PQ² Step 4: Solve for PQ PQ² = 169 - 25 PQ² = 144 PQ = √144 PQ = 12 cm Final Answer: The length of the tangent PQ is 12 cm.
• Example 2: Using Theorem 10.2 A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC. Step 1: Understand the Setup The quadrilateral's sides AB, BC, CD, and DA are tangents to the circle at points P, Q, R, and S, respectively. Step 2: Apply Theorem 10.2 According to Theorem 10.2, the lengths of tangents from an external point to a circle are equal. We apply this to each vertex of the quadrilateral: From point A: AP = AS (Equation 1) From point B: BP = BQ (Equation 2) From point C: CR = CQ (Equation 3) From point D: DR = DS (Equation 4) Step 3: Add the Equations Add all four equations together: (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ) Step 4: Group the Terms By looking at the figure, we can see that: AP + BP = AB CR + DR = CD AS + DS = AD BQ + CQ = BC Substituting these into the combined equation from Step 3, we get: AB + CD = AD + BC Final Answer: Hence, it is proved that for a quadrilateral circumscribing a circle, the sum of opposite sides is equal.

Exam Tips: Avoid These Common Mistakes!

When solving problems from the circles class 10 NCERT chapter, students often make a few common errors. Be careful to avoid them!

1. Confusing Tangent and Secant: A tangent touches the circle at one point, while a secant intersects it at two. Don't use tangent properties for a secant.
2. Incorrect Right Angle: Remember, the radius is perpendicular to the tangent at the point of contact. The angle will not be 90° anywhere else along the tangent.
3. Pythagoras Theorem Errors: Many problems use the Pythagoras theorem. Double-check your calculations and ensure you correctly identify the hypotenuse (it's always the side opposite the 90° angle, which is often the line from the center to the external point).
4. Misapplying Theorem 10.2: When a figure has multiple tangents, correctly identify the single external point from which two tangents are drawn. For example, in the quadrilateral problem, AP and BP are not equal; the equal pairs are AP and AS (from point A) and BP and BQ (from point B).

Practice Questions with Solutions

• Q: Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle. A: Step 1: Draw a diagram. Let the center be O. Let the larger circle be C1 and the smaller circle be C2. Let AB be the chord of C1 that is tangent to C2 at point P. Step 2: Join OP and OA. OP is the radius of the smaller circle (OP = 3 cm) and OA is the radius of the larger circle (OA = 5 cm). Since AB is a tangent to C2 at P, by Theorem 10.1, OP ⊥ AB. Step 3: In the right-angled triangle △OPA, apply Pythagoras theorem: OA² = OP² + AP². We get 5² = 3² + AP², which means 25 = 9 + AP². So, AP² = 16, and AP = 4 cm. Step 4: Since the perpendicular from the center to a chord bisects the chord, AP = PB. Therefore, the length of the chord AB = 2 AP = 2 4 = 8 cm. Final answer: The length of the chord is 8 cm.
• Q: From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the center is 25 cm. What is the radius of the circle? A: Step 1: Let the center of the circle be O and the point of tangency be T. We are given the length of the tangent QT = 24 cm and the distance from the center OQ = 25 cm. Step 2: According to Theorem 10.1, the radius is perpendicular to the tangent at the point of contact. So, OT ⊥ QT. This forms a right-angled triangle △OTQ. Step 3: Apply Pythagoras theorem. The hypotenuse is OQ. So, OQ² = OT² + QT². We need to find the radius, OT. Step 4: Substitute the given values: 25² = OT² + 24². This gives 625 = OT² + 576. So, OT² = 625 - 576 = 49. Therefore, OT = √49 = 7 cm. Final answer: The radius of the circle is 7 cm.
• Q: In the given figure, if TP and TQ are the two tangents to a circle with center O so that ∠POQ = 110°, find ∠PTQ. A: Step 1: Identify the given information. TP and TQ are tangents from point T. O is the center. ∠POQ = 110°. Step 2: Apply Theorem 10.1. The radius is perpendicular to the tangent at the point of contact. Therefore, OP ⊥ TP and OQ ⊥ TQ. This means ∠OPT = 90° and ∠OQT = 90°. Step 3: Consider the quadrilateral OPTQ. The sum of all angles in a quadrilateral is 360°. So, ∠POQ + ∠OPT + ∠OQT + ∠PTQ = 360°. Step 4: Substitute the known values: 110° + 90° + 90° + ∠PTQ = 360°. This simplifies to 290° + ∠PTQ = 360°. Therefore, ∠PTQ = 360° - 290° = 70°. Final answer: The angle ∠PTQ is 70°.
• Q: Prove that the tangents drawn at the ends of a diameter of a circle are parallel. A: Step 1: Let AB be the diameter of a circle with center O. Let PQ and RS be the tangents drawn at the endpoints A and B respectively. Step 2: Apply Theorem 10.1. The radius is perpendicular to the tangent at the point of contact. At point A, radius OA ⊥ tangent PQ. So, ∠OAP = 90°. At point B, radius OB ⊥ tangent RS. So, ∠OBR = 90°. Step 3: Now consider AB as a transversal line intersecting the lines PQ and RS. We have ∠OAP = 90° which is the same as ∠PAB = 90°. Also, ∠OBR = 90° which is the same as ∠ABS = 90° (as O lies on the line segment AB). Step 4: The angles ∠PAB and ∠ABS are alternate interior angles. Since ∠PAB = ∠ABS = 90°, the alternate interior angles are equal. When alternate interior angles are equal, the lines are parallel. Therefore, PQ || RS. Final answer: Hence, it is proved that the tangents drawn at the ends of a diameter are parallel.

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