Circles — Class 10 Maths Notes

Topics covered in Circles

1. 1.Introduction of Circles

   What is a Circle? A circle is a round shape that has no corners or edges. Imagine a coin or a dinner plate – these are everyday examples of circles.

   Key Elements of a Circle:

   • Center: The middle point of the circle.
   • Radius: The distance from the center to any point on the circle. It's like a straight line drawn from the middle of a cookie to its edge.
   • Diameter: A straight line that passes through the center and touches two points on the circle. It is exactly twice as long as the radius.
   • Circumference: The distance around the circle, like the boundary line of a park.

   How to Calculate?

   • Circumference: You can find it by multiplying the diameter with π (Pi, approximately 3.14). Formula: =×C=π×d.
   • Area: The space inside the circle. You calculate it using the radius and π. Formula: =×2A=π×r2.

   Real-Life Examples:

   • Wheels on vehicles are circles – this shape helps them roll smoothly.
   • Clocks often have circular faces, making it easier to display the time evenly around the center.

   Activities to Understand Circles:

   1. Take a round object like a jar lid and trace it on a paper. You've drawn a circle!
   2. Measure the diameter of a round object using a ruler and calculate its circumference.

   Where is it Used?

   • In Careers: Architects and engineers use circles to design wheels, gears, and arches.
   • In Industries: From manufacturing car tires to creating designs in graphic software, circles are everywhere.

2. 2.Tangent to a Circle

   What is a Tangent to a Circle?

   • A tangent to a circle is a straight line that touches the circle at exactly one point. This point is known as the point of tangency. Think of it like a straight road just touching the boundary of a roundabout and not crossing it.

   Key Features of a Tangent:

   • A tangent touches the circle at only one point.
   • It is perpendicular to the radius of the circle at the point of tangency.
   • A tangent does not intersect the circle; it only touches the outer edge.

   Real-Life Examples of Tangents:

   • The way a basketball touches the ground at only one point.
   • A mirror resting on the edge of a circular table.

   Simple Activities to Understand Tangents:

   1. Draw a circle on paper, then draw a straight line that touches the circle at only one point without crossing it. This line is a tangent.
   2. Place a ruler touching the edge of a round object like a bowl. The ruler represents a tangent.

   Use of Tangents in Real Life and Careers:

   • In careers like engineering and architecture, tangents are used for designing curved roads, wheels, or gears.
   • In industries like automotive, the concept of tangents helps in understanding tire contact with roads.

3. 3.Theorem 1 : The tangent at any point of a circle is perpendicular to the radius through the point of contact.

   To prove the theorem that the tangent at any point of a circle is perpendicular to the radius through the point of contact, we'll use a geometric approach. This theorem is a fundamental result in Euclidean geometry.

   Theorem: The tangent to a circle is perpendicular to the radius at the point of contact.

   Proof:

   1. Given: A circle with center O and a tangent XY at point P.

   2. To Prove: OP is perpendicular to XY at P.

   3. Construction: Extend line OP to any point Q on XY such that P is between O and Q. The line segment OQ is not a radius of the circle.

   4. Proof Method: By contradiction.

   5. Proof:

      • Suppose that OP is not perpendicular to XY. Then line OQ would form an angle with XY that is not 90 degrees at point P.

      • Since OQ is not a radius, but an extension of the radius beyond the circle, OQ is longer than OP. That is, >OQ>OP.

      • By the properties of circles, the shortest distance from the center of a circle to a tangent line is along the radius. Therefore, if OP were not perpendicular to XY, there would exist another point, Q, such that OQ is shorter than OP, which contradicts the definition of a radius being the shortest distance to the tangent.

      • Thus, our assumption that OP is not perpendicular to XY leads to a contradiction.

      • Therefore, OP must be perpendicular to XY.

   6. Conclusion: It follows that OP is perpendicular to XY at P, and our theorem is proved.

   This result is used in many real-world applications. For example, engineers may use this property when designing circular objects that interact with flat surfaces, such as gears and wheels. In architecture, this concept is utilized in creating rounded elements that meet flat surfaces, ensuring that the structures are stable and aesthetically pleasing.

4. 4.How many tangents can a circle have?

   A circle can have an infinite number of tangents. A tangent to a circle is a line that touches the circle at exactly one point. Since a circle is a continuous curve, every point on its circumference can serve as the point of contact for a tangent line. Thus, for every point on the circle's circumference, there can be a unique tangent line, leading to an infinite number of tangents.

   
   

   In practical terms, imagine drawing a straight line that just touches the outer edge of a circle without cutting through it. You can do this at any point along the circle's edge, meaning there's no limit to the number of tangents you can draw.

   
   

   In real-life applications, understanding tangents is important in various fields like engineering, architecture, and even in art for designing and visualizing geometric shapes and structures.

5. 5.A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is :

   Given:

   • Radius of the circle (OP) = 5 cm
   • OQ = 12 cm

   In a circle, the tangent at any point is perpendicular to the radius at the point of contact. Therefore, OP is perpendicular to PQ.

   We have a right-angled triangle OPQ with OP as the radius (5 cm) and OQ (12 cm) as the hypotenuse.

   To find the length of PQ, we use the Pythagorean theorem:

   2=2−2PQ2=OQ2−OP2

   Putting the given values:

   2=122−52PQ2=122−52 2=144−25PQ2=144−25 2=119PQ2=119

   So, =119PQ=119​ cm.

   Hence, the length of PQ is 119119​ cm.

6. 6.Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.

   
   
   

   We see a circle with a central point labeled O. There are two lines: line m, which is a tangent to the circle at point P, and line n, which is a secant intersecting the circle at points Q and R. The line m is parallel to the given line l, and similarly, the line n is also parallel to l.

   
   

   The line m touches the circle at exactly one point P, making it a tangent, and it's clear that the tangent forms a 90-degree angle with the radius OP of the circle, which is a characteristic property of tangents to a circle. The line n cuts through the circle at two distinct points, Q and R, which is the definition of a secant.

7. 7.Number of Tangents from a Point on a Circle

   When a point is on the circle, exactly one tangent can be drawn to the circle from that point. This tangent will touch the circle at only that point and will be perpendicular to the radius of the circle at the point of contact.

   For a point P on a circle with center O, the tangent at point P is a line that just touches the circle at P and nowhere else. If you tried to draw another line from point P that you also called a tangent, it would either enter the circle's interior (and thus not be a tangent) or overlap with the first tangent line.

   The concept of tangents to a circle from points in different positions relative to the circle:

   1. Case 1: Point P is inside the circle (I). As the activity suggests, any line through point P inside the circle intersects the circle at two points, so no tangent can be drawn from point P.

   2. Case 2: Point P is on the circle(II). In this case, there is exactly one tangent to the circle through point P. This tangent touches the circle at P and is perpendicular to the radius at that point.

   3. Case 3: Point P is outside the circle(III). From point P, two tangents 1PT1​ and 2PT2​ can be drawn to the circle, touching the circle at points 1T1​ and 2T2​ respectively. The lengths of these tangents from point P to the points of contact with the circle are called the lengths of the tangents.

   The activity hints at an important property of tangents to a circle from an external point: the lengths of the tangents from that point to the circle (here, 1PT1​ and 2PT2​) are equal. This is a fundamental result in circle geometry and is true for any point outside a circle; the tangents from the external point to the circle are always equal in length. This can be proven using various methods, such as using congruent triangles formed by the tangents and the radii to the points of tangency.

8. 8.Theoram 2 : The lengths of tangents drawn from an external point to a circle are equal.

   Theorem: The lengths of tangents drawn from an external point to a circle are equal.

   Proof:

   1. Given: A circle with center O and two tangents PQ and PR from an external point P.

   2. To Prove: PQ = PR.

   3. Construction: Draw the radii OQ and OR to the points of tangency Q and R.

   Proof Method: We will use the properties of tangents and the concept of congruent triangles.

   1. Tangents from a point to a circle are perpendicular to the radius at the point of tangency. Therefore, ⊥OQ⊥PQ and ⊥OR⊥PR.

   2. In triangles OPQ and OPR:

      • The radii OQ and OR are equal because all radii of a circle are equal (OQ = OR).
      • The line segments PQ and PR are both tangent to the circle from point P, and each is perpendicular to a radius at the point of tangency, forming right angles at Q and R (∠=∠=90∘∠OPQ=∠OPR=90∘).
      • OP is common to both triangles OPQ and OPR.

   3. Since both triangles have two sides equal and the angle between them is equal, by the Side-Angle-Side (SAS) postulate, the two triangles are congruent (△≅△△OPQ≅△OPR).

   4. By congruency, the lengths of the tangents PQ and PR are equal (PQ = PR).

   Conclusion: The lengths of tangents drawn from an external point to a circle are equal. This property is fundamental in geometry and has applications in various fields, such as engineering and physics, where it is used to solve problems involving circles and tangents.

9. 9.Prove that in two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.

   Two concentric circles (circles with the same center) with a common center O. There is a chord AB of the larger circle that touches the smaller circle at point P.

   To prove: The chord AB is bisected at the point of contact P, meaning AP = PB.

   Proof:

   Since the circles are concentric, they share the same center, O.

   1. Draw radii OA and OB to the endpoints of the chord AB. Since radii of the same circle are equal, OA = OB.

   2. Draw the radius OP to the point of tangency P. Since a radius drawn to a point of tangency is perpendicular to the tangent at the point, OP is perpendicular to AB. Therefore, ∠OAP and ∠OBP are right angles.

   3. We now have two right triangles, ΔOAP and ΔOBP, with:

      • OA = OB (radii of the same circle)
      • OP is common to both triangles
      • ∠OAP and ∠OBP are right angles

   By the Hypotenuse-Leg (HL) theorem (a specific case of congruency for right triangles), we can say that ΔOAP ≅ ΔOBP.

   If two triangles are congruent, all corresponding parts are equal. Therefore, AP = PB.

   Conclusion: The chord AB of the larger circle that touches the smaller circle is bisected at the point of contact P. This means that AP = PB.

10. 10.PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T (see Fig.). Find the length TP.

    A circle with radius 5 cm, a chord PQ that is 8 cm long, and tangents PT and QT that meet at point T outside the circle. To find the length of TP, we can apply the properties of tangents and the Pythagorean theorem.

    Let's denote R as the midpoint of the chord PQ. Since PQ is a chord of the circle, and OR is a radius that bisects it, OR is perpendicular to PQ. Therefore, triangle ORP is a right-angled triangle, with OR as one side and PR as the other side.

    Given that the length of PQ is 8 cm, PR, being half of PQ, is 4 cm. The radius OR is 5 cm. We can use the Pythagorean theorem to find the length of OP:

    2=2+2OP2=OR2+PR2

    2=52+42OP2=52+42

    2=25+16OP2=25+16

    2=41OP2=41

    =41OP=41​ cm

    Since TP is a tangent from point T to the circle at point P, and OP is the radius to the point of tangency, the length of the tangent TP is the same as OP (which is a property of tangents from an external point to a circle).

    Hence, the length of TP is 4141​ cm.

11. 11.Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

    A circle with center O and a diameter AB. There are tangents at points A and B, with the tangent at A extending to points P and Q, and the tangent at B extending to points R and S. We are to prove that these two tangents are parallel to each other.

    Proof:

    1. Given: A circle with center O and a diameter AB. Tangents at points A and B are extended to points P, Q, and R, S respectively.

    2. To Prove: Tangent at A (line PQ) is parallel to the tangent at B (line RS).

    3. Construction: Draw the radius OA to point A and the radius OB to point B. Since AB is a diameter, angle AOB forms a straight line and thus is 180 degrees.

    4. Proof Method: Use the property that a tangent to a circle is perpendicular to the radius at the point of tangency.

    • Since OA is a radius and PQ is a tangent at point A, angle OAP is 90 degrees ( ∠=90∘∠OAP=90∘).
    • Since OB is a radius and RS is a tangent at point B, angle OBS is also 90 degrees ( ∠=90∘∠OBS=90∘).

    1. Both tangents PQ and RS are perpendicular to the same line AB (which is the diameter of the circle). By definition, if two lines are perpendicular to the same line, they are parallel to each other.

    Conclusion: Thus, the tangent PQ at point A is parallel to the tangent RS at point B. The tangents drawn at the ends of a diameter of a circle are indeed parallel.

12. 12.Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

    Proof:

    1. Given: A circle with center O and a tangent line AB at point x. A line xy is drawn perpendicular to AB at point x.

    2. To Prove: Line xy passes through the center O.

    3. Construction: Draw the line Ox from point O (the center of the circle) to point x (the point of tangency).

    4. Proof Steps:

       • By the definition of a tangent, a tangent to a circle is a line that touches the circle at exactly one point and is perpendicular to the radius of the circle at that point.
       • In the figure, AB is the tangent that touches the circle at point x.
       • Ox is the radius of the circle that ends at point x, the point of tangency.
       • According to the property of tangents, radius Ox is perpendicular to the tangent line AB at the point x. This means that the angle formed by Ox and AB is 90 degrees (∠=90∘∠OxAB=90∘).
       • The line xy is also drawn perpendicular to AB at point x. This means ∠=90∘∠xyAB=90∘ as well.
       • In Euclidean geometry, if two lines (Ox and xy) are both perpendicular to the same line (AB) at the same point (x), they must lie on the same line. In other words, Ox and xy cannot be two separate lines because that would mean there would be two different perpendiculars from the same point to the same line, which is not possible.
       • Therefore, line xy and line Ox must be the same line. Since Ox passes through the center O of the circle, line xy must also pass through O.

    Conclusion: This geometric reasoning concludes that the perpendicular line xy at the point of tangency to the tangent AB of the circle must pass through the center O of the circle. This is based on the unique property that a tangent to a circle is perpendicular to the radius at the point of tangency.

13. 13.The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.

    In the given problem, we have:

    • OA = 5 cm (the distance from point A to the center O of the circle)
    • AT = 4 cm (the length of the tangent from point A to the circle)

    Since the tangent to a circle is perpendicular to the radius at the point of contact (by the tangent-radius theorem), we have a right-angled triangle OAT with OT as the radius, AT as the tangent, and OA as the line from the center to the point A.

    Applying the Pythagorean theorem to the right-angled triangle OAT, we get:

    2=2+2OA2=OT2+AT2

    Given that =5OA=5 cm and =4AT=4 cm, we can solve for OT (the radius):

    2=2−2OT2=OA2−AT2 2=52−42OT2=52−42 2=25−16OT2=25−16 2=9OT2=9

    Taking the square root of both sides:

    =9OT=9​ =3OT=3 cm

    Therefore, the radius of the circle is 3 cm.

14. 14.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.

    
    

    In the given problem, we have two concentric circles (circles with the same center) where the radius of the larger circle is 5 cm and the radius of the smaller circle is 3 cm. We need to find the length of the chord AB of the larger circle which touches the smaller circle.

    The chord AB of the larger circle that touches the smaller circle forms a right angle with the radius of the smaller circle at the point of contact P (since the tangent at any point of a circle is perpendicular to the radius at the point of contact).

    Here's how you can find the length of the chord AB:

    1. The radius OP of the smaller circle extends to the center O of the larger circle and is given as 3 cm.

    2. The radius OA of the larger circle is given as 5 cm.

    3. Since OP is perpendicular to AB and OA is the radius that extends to the edge of the chord, triangle OPA is a right triangle.

    4. Using the Pythagorean theorem, the length of AP, which is half of the chord AB, can be found by:

       2=2+2OA2=OP2+AP2

       2=2−2AP2=OA2−OP2

       2=52−32AP2=52−32

       2=25−9AP2=25−9

       2=16AP2=16

       =16AP=16​

       =4AP=4 cm

    5. Since AP is half of the chord AB, the full length of the chord AB is 2×2×AP.

    6. Therefore, the length of the chord AB is 2×42×4 cm = 88 cm.

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

15. 15.A quadrilateral ABCD is drawn to circumscribe a circle (see Fig.) Prove that AB + CD = AD + BC

    Given a quadrilateral ABCD that circumscribes a circle, we will label the points of tangency on the circle as,,,P,Q,R, and S such that P is on AB, Q is on BC, R is on CD, and S is on DA.

    To Prove: +=+AB+CD=AD+BC

    Proof:

    1. Let the lengths of the tangents from A to points P and S be x and y, respectively. So, ==AP=AS=x and ==DS=DP=y.

    2. Let the lengths of the tangents from B to points P and Q be z. So, ==BP=BQ=z.

    3. Let the lengths of the tangents from C to points Q and R be w. So, ==CQ=CR=w.

    4. Let the lengths of the tangents from D to points R and S be v. So, ==DR=DS=v.

    Now, the lengths of the sides of the quadrilateral can be expressed as follows:

    =+=+AB=AP+BP=x+z

    =+=+BC=BQ+CQ=z+w

    =+=+CD=CR+DR=w+v

    =+=+DA=DS+AS=v+x

    Adding AB and CD, we get:

    +=(+)+(+)AB+CD=(x+z)+(w+v)

    Adding AD and BC, we get:

    +=(+)+(+)AD+BC=(x+v)+(z+w)

    By rearranging the terms, we can see that:

    +=+++AB+CD=x+z+w+v

    +=+++AD+BC=x+v+z+w

    Hence, both expressions are equal:

    +=+AB+CD=AD+BC

    This completes the proof.

    This property is important in the study of cyclic quadrilaterals and is useful in various mathematical problems and proofs, including those related to the properties of tangents and circumscribed polygons.

16. 16.In Given figure XY and X'Y' are two parallel tangents to a circle with centre O and and another tangent AB with point of contact C interesting XY at A and X'Y' at B prove that ∠AOB = 90°.

    Given information from the diagram, where
    

    XY and ′′X′Y′ are parallel tangents to a circle with center O, and another tangent AB touches the circle at point C, intersecting XY at A and ′′X′Y′ at B, we are asked to prove that the angle ∠=90°∠AOB=90°.

    To prove this, we will use the properties of tangents and angles:

    1. Tangents from a point outside a circle are equal in length.
    2. The radius is perpendicular to the tangent at the point of contact.

    Proof:

    • Since XY and ′′X′Y′ are parallel tangents, and AB is a tangent that intersects XY at A and ′′X′Y′ at B, the line segment AB is a transversal to the parallel lines XY and ′′X′Y′.

    • The radius OC is perpendicular to the tangent AB at the point of contact C. This means that ∠=∠=90°∠OCA=∠OCB=90°.

    • Points A, O, and B lie on a straight line, with O being the midpoint, because OA and OB are radii of the two sectors created by the tangent AB. Since OC is perpendicular to AB, and O lies on the line segment AB, it follows that OC bisects AB at right angles.

    • Therefore, triangle AOC is congruent to triangle BOC by the hypotenuse-leg (HL) theorem since OC is a common side, =OA=OB (radii of the same circle), and ∠=∠=90°∠OCA=∠OCB=90°.

    • Since ∠∠OCA and ∠∠OCB are right angles, the angle formed by OA and OB at point O, which is ∠∠AOB, must also be a right angle. This is because the angles around point O must add up to 360°360°, and ∠+∠=180°∠AOC+∠COB=180°, leaving ∠∠AOB to be 180°−180°=0°180°−180°=0°, which means ∠∠AOB is a straight line and thus ∠=90°∠AOB=90°.

    Hence, we have proven that ∠=90°∠AOB=90°. This is a characteristic property of a circle and its tangents, and such properties are often used in geometric proofs and constructions.

17. 17.Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre

    Proof:
    

    • Given: A circle with center O and point P outside the circle. Tangents PA and PB are drawn from P to the circle at points A and B, respectively.
    • To prove: ∠APB + ∠AOB = 180°
      
      
      

    Steps:

    1. Draw radius OA.
    2. Since PA is a tangent to the circle at A, we know that ∠OAP = 90°.
    3. Similarly, since PB is a tangent to the circle at B, we know that ∠OBP = 90°.
    4. Therefore, ∠OAP + ∠OBP = 90° + 90° = 180°.
    5. Since ∠APB and ∠AOB are supplementary angles, we have ∠APB + ∠AOB = 180°.

    Note: The above proof is based on the fact that a tangent to a circle is perpendicular to the radius at the point of tangency. This is a basic property of circles that can be proven using similar triangles.

18. 18.Prove that the parallelogram circumscribing a circle is a rhombus.

    To prove that a parallelogram circumscribing a circle is a rhombus, we can use the property that the tangents from a single external point to a circle are equal in length.
    

    Proof:

    1. In the given figure, the circle is touching the parallelogram ABCD at points P, Q, R, and S.

    2. 1. AP = AS, BP = BQ, CQ = CR, and DR = DS because the lengths of tangents from an external point to a circle are equal.

    3. 2. Since ABCD is a parallelogram, opposite sides are equal (AB = CD and AD = BC).

    4. 3. Therefore, we have AP + BP = AS + DS and BQ + CQ = DR + CR.

    5. 4. Simplifying, we get AB = CD and AD = BC since AP = AS, BP = BQ, CQ = CR, and DR = DS.

    6. 5. This means all four sides of the parallelogram are equal (AB = BC = CD = DA).

    7. A parallelogram with all sides equal is defined as a rhombus.

    Therefore, the parallelogram circumscribing a circle must be a rhombus.

19. 19.A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Fig.). Find the sides AB and AC.

    To find the sides AB and AC of the triangle ABC which circumscribes a circle with radius 4 cm, and given that BD = 8 cm and DC = 6 cm, we will use the property of tangents to a circle from an external point being equal.
    
    
    

    
    

    Here is a step-by-step explanation:

    
    

    1. Let the points where the circle touches sides AB and AC be E and F, respectively.

    2. Since AE and AF are tangents to the circle from point A, we have AE = AF.

    3. Similarly, since BD and DC are tangents from points B and C, we have BE = BD and CF = DC.

    4. Therefore, BE = 8 cm and CF = 6 cm.

    
    

    Now, using these tangents:

    
    

    - AB = AE + BE

    - AC = AF + CF

    
    

    Because AE = AF, we can find the length of AE or AF by subtracting BE or CF from the length of BC, since AE + BE + CF = BC.

    
    

    Let's calculate BC first:

    
    

    BC = BD + DC

    BC = 8 cm + 6 cm

    BC = 14 cm

    
    

    Now, to find AE (or AF):

    
    

    - AB = AE + BE

    - AC = AE + CF

    
    

    Since AE = AF and BE = BD and CF = DC, we can write:

    
    

    - AB = AE + 8 cm

    - AC = AE + 6 cm

    
    

    To find AE or AF, we use the fact that the sum of the tangents from one point (A) to the circle is equal to the tangent from the other point (D), hence AE + AF = BD + DC. Since AE = AF, this means:

    
    

    2 * AE = BD + DC

    2 * AE = 8 cm + 6 cm

    2 * AE = 14 cm

    AE = 7 cm

    
    

    So, now we can find AB and AC:

    
    

    AB = AE + BE = 7 cm + 8 cm = 15 cm

    AC = AE + CF = 7 cm + 6 cm = 13 cm

    
    

    Therefore, AB is 15 cm and AC is 13 cm.

20. 20.Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

    To prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of the circle, we need to consider the angles that the sides of the quadrilateral make at the center of the circle.
    

    Given a circumscribed quadrilateral ABCD, with the circle touching the quadrilateral at points P, Q, R, and S, we have the following angles at the center O of the circle created by the sides of the quadrilateral:

    • ∠∠AOB created by side AB
    • ∠∠BOC created by side BC
    • ∠∠COD created by side CD
    • ∠∠DOA created by side DA

    To prove that the sum of the angles subtended by any two opposite sides is supplementary (sums to 180 degrees), we will use the property of tangents and the angles they subtend at the center of the circle.

    Proof:

    1. Tangents to a circle from an external point are equal in length. Hence, AP = AS, BP = BQ, CR = CQ, and DR = DS.

    2. Since AP and AS are equal, they subtend equal angles at the center O, meaning ∠=∠∠AOP=∠AOS.

    3. Similarly, ∠=∠∠BOP=∠BOQ, ∠=∠∠COQ=∠COR, and ∠=∠∠DOS=∠DOR.

    4. The angles ∠,∠,∠,∠AOB,∠BOC,∠COD, and ∠∠DOA are the external angles for these equal angles at the center O.

    5. The sum of the angles around point O is 360 degrees. Therefore, we have: ∠+∠+∠+∠=360∘∠AOB+∠BOC+∠COD+∠DOA=360∘

    6. The sum of the angles at the center O subtended by two opposite sides (for example, AB and CD) is: ∠+∠∠AOB+∠COD

    7. Since ∠∠AOB and ∠∠COD are exterior angles for adjacent angles ∠,∠∠AOP,∠AOS (and similarly for ∠,∠∠COQ,∠COR), we can say: ∠=∠+∠∠AOB=∠AOP+∠AOS ∠=∠+∠∠COD=∠COQ+∠COR

    8. We can substitute the expressions for ∠∠AOB and ∠∠COD back into the equation for the sum of the angles around point O: (∠+∠)+(∠+∠)+∠+∠=360∘(∠AOP+∠AOS)+(∠COQ+∠COR)+∠BOC+∠DOA=360∘

    9. Since ∠+∠+∠+∠∠AOP+∠AOS+∠COQ+∠COR represent all the angles around point O, and the sum of all angles around point O is 360 degrees, the sum of the remaining two angles ∠∠BOC and ∠∠DOA must be 180 degrees.

    10. Therefore, ∠+∠=180∘∠BOC+∠DOA=180∘, and by the same argument, ∠+∠=180∘∠AOB+∠COD=180∘.

    Hence, the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of the circle.

21. 21.Quick Revision

    1. Introduction to Circles: A circle is a round shape where every point on its edge is the same distance from the center. This distance is called the radius. The circle also has a diameter, which is a line going straight across the circle, passing through the center, and it's twice the length of the radius.

    2. Tangent to a Circle: A tangent is a straight line that touches a circle at exactly one point. It never enters the circle's interior and it's at a right angle to the radius at the point of contact.

    3. Theorem 1: The theorem states that if you draw a tangent to a circle, then it will be perpendicular (form a 90-degree angle) to the radius at the point where it touches the circle.

    4. How Many Tangents Can a Circle Have? A circle can have an infinite number of tangents because there are infinite points along the circle's circumference where a tangent can touch.

    5. Number of Tangents from a Point on a Circle: From any point on a circle, there will be exactly one tangent that you can draw. This is because only one line can touch the circle at that point and still be straight.

    6. Theorem 2: This theorem says that if you have a point outside of a circle and you draw two tangents from that point to the circle, both tangents will be the same length.

    7. Proving that the Parallelogram Circumscribing a Circle is a Rhombus: To prove this, you would show that all sides of the parallelogram are equal, which is a property of a rhombus. This happens because the tangents from a point outside the circle to the circle are equal, and in a parallelogram circumscribing a circle, opposite sides are made up of two such tangents each.

    Important Formulas:

    • The length of a tangent from a point A outside the circle to the point of contact T on the circle is given by =2−2AT=PA2−r2​ where P is the center of the circle and r is the radius.

    To use this formula, measure the distance from the point outside the circle to the center of the circle (that's PA), and subtract the square of the radius. Then take the square root to find the length of the tangent.