Circles (Concepts, properties and CAT questions)
Monday, July 20th, 2020
Concepts of a circle are very important for CAT examinations. There are lots of properties to understand and some formulas to remember. Several direct and sometimes indirect questions are asked from concepts of a circle in CAT exams. Also, some of the geometry questions cannot be solved without having a proper understanding of CAT concepts. Let’s take a look at some of the concepts and CAT questions related to circles.
1. Arc:
An arc is just a part of a circle. An arc can be measured in degrees.
In the above diagram, the part of the circle from B to C forms an arc. Also, arc BC is equal to the angle BOC that is 45°.
1.1 Important Terms of Arcs:
(i) Measure of an Arc
1) The measure of a semi-circle is 1800.
2) The measure of a minor arc is the measure of its central angle. In the figure, BC is the minor arc with the measure 450
3) The measure of a major arc is 3600-(measure of corresponding minor arc)
m (arc BAC)=3600-m(Arc BDC) = 3600 – 450 = 3150
2. Tangent
A tangent is a line that touches a circle at only one point. A tangent is perpendicular to the radius at the point of contact.
In the above diagram, the line containing the points B and C is a tangent to the circle. It touches the circle at point B and is perpendicular to the radius OB. BC is perpendicular to OB.
3. Chord
A chord is also a line segment with both endpoints on the circle, but it may not pass through the center of the circle.
4. Secant
It is a line which intersects the circle in two distinct points. See the below figure for a definition of chord and segment
Circumference and Area of the circle
Circumference:
The circumference of a circle is the distance around a circle.
The formula for the circumference of a circle is C = πd (π =3.142)
Or C = 2Ï€r, where C is the circumference, d is the diameter and r is the radius.
Area
The area of a circle is the region enclosed by the circle.
It is given by the formula: A = πr2 where A is the area and r is the radius.
Types of circles:
Concentric Circles: Circles lying in the same plane with a common center are called concentric circles
Tangent Circles: Circles lying in the same plane and having only one point in common are called tangent circles
One and only one circle passes through three given non-collinear points. An infinite number of circles pass through two given points.
Important Properties of Chords, Tangents and Secants:
PROPERTY 1:
The perpendicular from the center of a circle to a chord of the circle bisects the chord.
Conversely, the line joining the center of the circle and the midpoint of the chord is perpendicular to the chord.
If OM =AB, then AM=MB and If AM=MB, Â then OM=AB
PROPERTY 2:
Equal chords of a circle are equidistant from the center. Conversely, two chords of a circle that are equidistant from the center and are making the same angle with the line drawn from the center of the circle are equal.
If AB=CD, then OM=ON and if OM=ON, then AB=CD
PROPERTY 3:
In a circle, equal chords subtend equal angles at the center
Conversely, chords which subtend equal angles at the center of the same circle, are equal
If AM=CD, then m∠COD=m∠AOB and If m∠COD= m∠COD, then AB=CD
PROPERTY 4:
Tangent Perpendicularity Theorem:
The tangent at any point of a circle and the radius through that point are perpendicular to each other. If O is the center of the circle, A is the point of contact of the tangent X, then OA is perpendicular to X
Given a point on a circle, there is one and only one tangent to the circle passing through that point
From an exterior point of the circle, exactly two tangents can be drawn onto the circle.
PROPERTY 5:
The lengths of two tangent segments, from the exterior point to the circle, are equal. AC=BC.
This can go on to form a kite as follows
PROPERTY 6:
Angles inscribed in the same arc are equal m∠AOB=m∠ACB as they are inscribed in the same arc ADB
PROPERTY 7:
Equal arcs of a circle have equal chords. Conversely, equal chords of a circle have their corresponding arcs equal.
If m (arc AXB) =m (arc CYD), then AB=CD and If AB=CD, then m (arc AXB) = m (arc CYD)
PROPERTY 8:
Inscribed angle theorem:-The measure of an inscribed angle is half the measure of its intercepted arc. m ∠ACB= 1/2m(arc ADB)=1/2 m∠AOB
PROPERTY 9:
The angle that the tangent to the circle makes with a chord drawn from a point of contact is equal to the angle subtended by that chord in the alternate segment of the circle
In the following figure, x1=x2=x3
PROPERTY 10:
Opposite angles in a cyclic quadrilateral add up to 180o.
In the following figure, a+b=1800
PROPERTY 11:
If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the segment, then the four points lie on the same circle.
Points A, B, C, D lie on the same circle; i.e. they are concyclic
PROPERTY 12:
If two secants intersect in the exterior of the circle, the angle so formed is equal to half the difference of the measures of the arcs intercepted by them.
m(AOC)=1/2[m(arc AC)-m(arc BD)]
Also, AO x BO = CO x DO
PROPERTY 13:
If two secants intersect in the interior of the circle, the angle so formed is equal to half the sum of the measures of the arcs intercepted by them.
AOC=1/2[m (arc AC) +m (arc BD)] Also, AO x BO = CO x DO
Note: –Â O need not be the origin
PROPERTY 14: Common Tangents
For the two circles with centers A and B, PQ and RS are the direct common tangents and CD and EF are the transverse common tangents (Only two of both direct common and transverse common tangents are possible).
Where r1 and r2 are the radii of the two circles Length of Direct common tangent = v((Distance between centers)2 – (r1-r2 )2 )
Length of transverse common tangent = v ((Distance between centers) 2 – (r1+r2)2 )
Now, Let us take a look at few examples from the concepts of circles:
Examples:
Example 1:Â Â Five concentric circles are drawn whose circumferences are in the ratio 2:3:7:9:13. If a line is drawn joining the center to the outermost point on the circle, what is the ratio of the length of the line to the radius of the second innermost circle?
Solution: Since the circles are concentric, the line joining the center of the circle to the point on the outermost circle will be the radius of the outermost circle.
Now, Let the length of the radius of inner circle be 2x.
Thus, length of the radius of second circle would be 3x.
Similarly length of the radius of the outermost circle would be 13x.
Thus required ratio = 13x/2x = 13/2
Example 2: A palm tree swings with the tree in such a manner that the angle covered by its tree is 14 degrees. If the topmost portion of the tree covers a distance of 33 meters, find the length of the tree?
Solution: Let the length of the tree be OA. As per the question, AB is the distance covered by the tree. Thus length of arc AB = 33 meters. Also, ∠AOB = 14 degree.
We know that for a circle with radius r (OA = length of the tree) and angle θ at the center, measure of arc is given by 2πrθ/360
In our case, r= OA and θ = 14 degrees
Thus, 2 * (22/7) * r * (14/360) = 33 meters
Solving, we get r = 135 meters
Example 3:Â A circle of maximum possible size is cut from a square sheet. Subsequently, a square of maximum possible size is cut from the resultant circle. What will be area of the final square?
a.) 75% of the size of the original square
b.) 50% of the size of the original square
c.) 75% of the size of the circle
d.) 25% of the size of the original square
Solution: Let the side of the original square be ‘a’ units. Therefore, the area of the original square = a2Â units.
Concept 1: The diameter of the circle of maximum possible dimension that is cut from the square will be the side of the square. So, its diameter will be ‘a’ units.
Concept 2: The diagonal of the square of maximum possible dimension that can be cut from the circle will be the diameter of the circle. So, the diagonal of the final square will be ‘a’ units.
If the diagonal of the final square is ‘a’ units, then its side = a/√2 unis
If the side of the final square is a/√2 units, then its area = a2 / 2 units
Therefore, the area of the new square will be 50% of the area of the original square.
Example 4: Find the area of the sector covered by the hour hand after it has moved through 3 hours if the length of the hour hand is 7cm.
- 77 sq.cm
- 38.5 sq.cm
- 35 sq.cm
- 70 sq.cm
Solution: If the hour hand moved through 12 hours, it subtends 3600Â at the centre of the clock.
i.e., in 12 hours it makes 3600.
Therefore, in 3 hours it makes (3/12) x 360 = 900
The length of the hour hand is the radius of the circle.
We know that length of the hour hand = 7 cm. i.e., the radius of the circle is 7 cm.
Therefore, area of sector = (90/360) x π r2 = 1/4 x 22/7 x 7 x 7 = 38.5 sq.cm
The correct answer is Choice (2)
Example 5: What is the circumradius of a triangle whose sides are 7, 24 and 25 respectively?
- 18
- 12.5
- 12
- 14
Solution: 7, 24, 25 is a Pythagorean triplet. Therefore, the given triangle is a right angled triangle.
In a right-angle triangle, the circumradius measured half the hypotenuse. The hypotenuse of the give right angle triangle is 25. Therefore, its circumradius is 12.5 units.
Summary: Â In this post, we have seen some of the important properties of circles and some sample CAT questions. But, there are a huge variety of CAT questions other than the ones which we have seen above. We will go through some of the more CAT questions related to circles in the next post.
CAT Questions related to Quantitative Aptitude – Geometry
All questions from CAT Exam Quantitative Aptitude – Geometry
Quantitative Aptitude – Geometry – Triangles – Q1: Let P be an interior point of a right-angled isosceles triangle ABC with hypotenuse AB.
Quantitative Aptitude – Geometry – Triangles – Q2: Let ABC be a right-angled triangle with BC as the hypotenuse. Lengths of AB and AC are 15 km and 20 km, respectively.
Quantitative Aptitude – Geometry – Triangles – Q3: From a triangle ABC with sides of lengths 40 ft, 25 ft and 35 ft, a triangular portion GBC is cut off where G is the centroid of ABC.
Quantitative Aptitude – Geometry – Circles – Q1: ABCD is a quadrilateral inscribed in a circle with centre O. If ∠COD = 120 degrees and ∠BAC = 30 degrees
Quantitative Aptitude – Geometry – Circles – Q2: Let ABC be a right-angled isosceles triangle with hypotenuse BC. Let BQC be a semi-circle, away from A, with diameter BC.
Quantitative Aptitude – Geometry – Coordinate – Q1: The points (2, 5) and (6, 3) are two end points of a diagonal of a rectangle. If the other diagonal has the equation y = 3x + c, then c is
Quantitative Aptitude – Geometry – Coordinate – Q2: The shortest distance of the point (½, 1) from the curve y = |x -1| + |x + 1| is
Quantitative Aptitude – Geometry – Mensuration – Q1: The base of a vertical pillar with uniform cross section is a trapezium whose parallel sides are of lengths 10 cm and 20 cm
Quantitative Aptitude – Geometry – Mensuration – Q2: A ball of diameter 4 cm is kept on top of a hollow cylinder standing vertically.
Quantitative Aptitude – Geometry – Mensuration – Q3: A solid metallic cube is melted to form five solid cubes whose volumes are in the ratio 1 : 1 : 8: 27: 27.
Quantitative Aptitude – Geometry – Polygon – Q1: Let ABCDEF be a regular hexagon with each side of length 1 cm. The area (in sq cm) of a square with AC as one side is
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ratio answer to the first question is 13:3 , wrong answers are given how do u think one willl buy course from you.