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A circle is a two-dimensional geometrical figure formed by a set of all points at equidistant from a fixed point on the same plane. The perimeter of the circle is also called the circumference of a circle and an area of a circle is the region circumscribed by it.
Circle
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A circle in geometry is a unique kind of ellipse whose eccentricity is zero and the two foci are coincident. A circle is also known as the locus of points that are drawn equidistant from a point called “center”. The formula of a circle in a plane has been mentioned below:
(x – h)2 + (y – k)2 = r2
In the above formula for a circle,
(x, y) denote the coordinate points
(h, k) denotes the coordinates of the circle’s center.
A circle has a closed 2 dimensional (2D) figure which is measured in terms of radius. Circle separates the plane into two regions called the exterior and the interior region.
Every shape in geometry has an area and perimeter.
- Area
The area of a circle is the region bounded by the circle in a 2D plane.
Area of the circle: πr2
- Perimeter
The circle’s perimeter is simply its circumference. It is the distance around the circle.
The perimeter of the circle: πd = 2πr
The standard value of pie is considered as π = 3.1415
Explanation:
The concepts in the chapter Circle have been mentioned below:
- Tangents
A line which touches the circle at exactly one single point is known as the tangent to a circle. For every point on a circle, there is always a unique tangent passing through it. In the figure below, PQ is a tangent that touches the circle at point A.
Tangents
- Secant
A line that has two points in common with the circles is called a secant to circle. A secant cuts the circles at two points, forming a chord of the circle.
The chapter includes few important theorems which are:
- Tangent perpendicular to the radius of the circle at the point of contact
- Number of tangents drawn from a point that is given.
- Lengths of a tangent
- Lengths of tangents drawn from an external point
Important Questions for Circle
The questions from this chapter in the class 10 board exam can be short answer or long answer type questions. Some important questions from the chapter Circle along with their solutions have been mentioned below which will help you in preparing for the exams. The students are required to practice these questions to get familiar with the type of questions that might be asked in the examination.
Short answer type questions
Ques 1: A circle’s radius is 10 cm and its chord subtends a right angle at the circle’s center. Find out the length of the chord. (1 mark)
Answer: By applying Pythagoras theorem,
AB2 = OA2 + OB2
AB2 = 102 + 102
AB2 = 2(10)2
AB = 10√2 cm
Ques 2: PQR is an isosceles triangle and QR is a tangent to the circle having center X. Find ∠PQR. (1 mark)
Answer: Since the PQR is an isosceles triangle with tangent QR, the ∠PQR calculated will be 45°.
Ques 3: In the figure given below, PB and PA are two tangents to a circle having O as a center. If it is given that APB = 60°, find out the OAB. (2 marks)
Answer: From the given figure,
∠1 = ∠2
∠1 + ∠2 + ∠APB = 180°
∠1 + ∠1 + 60° = 180°
2∠1 = 180° – 60° = 120°
∠1 = 120°/2 = 60°
∠1 + ∠OAB = 90°
60° +∠OAB = 90°
∠OAB = 90° – 60° = 30°
Ques 4: Find the radius of the circle inscribed in the right triangle ABC where BC=12 cm, AB=5 cm, and right angle at B. (2 marks)
Answer: Given:
AB = 5 cm, BC = 12 cm
Using Pythagoras theorem,
AC2=AB2+BC2
= 52+122
= 25+144
= 169
AC=13.
We know that two tangents drawn to a circle from the same point that is exterior to the circle are of equal lengths.
So, AM=AQ=a
Similarly MB=BP=b and PC=CQ=c
We know
AB=a+b=5
BC=b+c=12 and
AC=a+c=13
Solving simultaneously we get a=3,b=2 and c=10
We also know that the tangent is perpendicular to the radius
Thus OMBP is a square with side b.
Hence, the length of the radius of the circle inscribed in the right angled triangle is 2cm.
Ques 5: What are tangent and Secant? (2 marks)
Answer: The line that meets the circle at only one point is called the Tangent to a circle.
The line which meets the circle at two points while intersecting the circle is called a Secant. The two points are always distant.
Long Answer Questions
Ques 1: Prove that the tangents that are drawn at the ends of a circle’s diameter are parallel. (2 marks)
Answer: At first, draw a circle, now connect two points P and Q such that PQ becomes the diameter of the circle having center O. Now, draw two tangents AB and CD at points P and Q respectively.
Now, both the radii i.e. PO and OQ are perpendicular to the tangents.
So now, OQ is perpendicular to CD
and OP perpendicular to AB
Therefore, ∠OPA = ∠OPB = ∠OQC = ∠OQD = 90°
From the above figure,
∠OQC and ∠OPB are alternate interior angles.
Also,
∠OQC = ∠OPB and ∠OQD = ∠OPA
Therefore, it can be concluded that line AB and the line CD will be parallel to each other.
Hence proved.
Ques 2: If in a circle, point O is the center point, PQ is a chord and PT is the tangent at point P and POQ measures 70° then calculate the TP. (2 marks)
Answer: Given that, angle POQ = 70°
From the triangle PQO, we can say that
1 = 2 this means,
1 + 2 + 70° = 180°
21 = 110°
This implies, 1 = 55°
1 + ∠TPQ = 90°
After substituting the values,
50° + TPQ = 90°
This implies, TPQ = 90° - 55°
= 35°
Hence, the angle TP is= 35°
Ques 3: We have 2 concentric circles that have a radii of 5cm and 3cm. You have to calculate the length of the chord of the larger circle touching the smaller circle. (3 marks)
Answer: From the given diagram,
AB is the tangent to smaller circle
Therefore, OP ⊥ AB
Applying Pythagoras theorem on triangle OPA,
OA2 = AP2 + OP2
=> 52 = AP2 + 32
=> AP2 = 25 – 9 = 16
=> AP = 4
OP ⊥ AB
Since the perpendicular line from the circle’s center bisects the chord;
AP will be equal to PB
So, AB = 2AP
= 2 × 4
= 8cm
Therefore, the larger circle will have an 8 cm long chord.
Ques 4: In a circle O is the center, AB and BC are two tangents to the circle in a way that ∠BAC =40°. Find out the ∠BOC. (2 marks)
Answer: Given that AB and BC are the tangents,
Therefore, ∠AOB = ∠AOC = 90°
In ABOC,
∠AOC + ∠AOB + ∠ABC + ∠BOC= 360°
90° + 90° + 40° + ∠BOC = 36°
∠BOC = 360° - 220°
= 140°
Hence, ∠BOC = 140°
Ques 5: We have a triangle ABC where, a = BC, b = CA, and AB = c. A circle touches the sides AB, BC, CA, of the triangle at F, D, and E respectively. You have to prove that BD = s – b. (3 marks)
Answer: Given-
A triangle ABC where a = BC
b = CA
and AB = c
Also, a circle is inscribed inside which touches the sides AB, BC, CA, of the triangle at F, D, and E respectively and s is the semi perimeter of the triangle
To Prove: BD = s – b
Proof:
According to the question,
We have,
Semi Perimeter = s
Perimeter = 2s
2s = AB + BC + AC….. Eq 1
As we know,
Tangents that are drawn to a circle from an external point are equal
So we have
AF = AE ….. Eq 2
BF = BD ….. Eq 3
CD = CE ….. Eq 4
Adding equation 2,3 and 4 we will get,
AF + CD + BF = AE + CE + BD
AB + CD = AC + BD
Adding BD on both sides,
AB + CD + BD = AC + BD + BD
AB + BC – AC = 2BD
AB + AC – AC + BC – AC = 2BD
2s – 2AC = 2BD (From equation 1)
2BD = 2s–2b (because AC = b)
BD = s – b
Hence proved.
Ques 6: A circle with center O has two tangents PA and PB making ∠APB = 50°. Find the ∠OAB. (2 marks)
Answer: PA = PB (Since, tangents drawn from external point are equal)
∠OAP = ∠OBP = 90°
∠OAB = ∠OBA (Angles opposite equal sides)
∠AOB + ∠OAP + ∠OBP + ∠APB = 360° (According to the Quadratic rule)
∠AOB + 90° + 90° + 50° = 360°
∠AOB = 360°–230°
= 130°
∠AOB + ∠OBA + ∠OAB = 180°
130° + 2∠OAB = 180°
2∠OAB = 50°
Thus, ∠OAB = 25°
Ques 7: In a given circle AP, BC, and AQ are the tangents where, AB = 5cm, AC = 6 cm, and BC = 4 cm. Find out the length of AP. (3 marks)
Answer: Since we know that length of tangents drawn on cine from an external point are equal
So, BP = BD .....(i)
CQ = CD .....(ii)
AP = AQ .....(iii)
AP = AB + BP
⇒AP = AB + BD .....(iv)
AQ = AC + CQ
⇒AQ = AC + CD .....(v)
Adding equation (iv) and (v)
AP + AQ = AB + BD + AC + CD
⇒ AP + AP = AB + BC + AC [By using equation (iii)]
⇒ 2AP = 11 + BC
⇒ 2AP = 11 + 4
⇒ 2AP = 15
AP = 7.5cm
Hence, AP = 7.5 cm
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