Develop Concepts with RS Aggarwal Class 11 Solutions Chapter-21 Circle

Circle is a short but important chapter for the students of Class 11. It forms some really important fundamentals that are essential to be known for the next chapters. Therefore, having a really strong foundation for this chapter is the key to good scores in the board and competitive exams. Solve the problems in your school textbook first and then move on to more conceptual problems like the ones in RS Aggarwal. Take advantage of the free PDF from Vedantu and make the most out of it by adding it to your study routine. 

Students of Class 11 must also remember that it is essential to clear their doubts immediately. Vedantu also helps in clearing doubts for students easily and offers them a great chance to work towards their goals. 

Short Description of Class 11 Maths Circle

Class 11 Maths RS Aggarwal Solutions is explained in simple steps here. These explanations are easy to understand and clear all doubts at once.  Class 11 Maths RS Aggarwal Solutions provides for easy exam preparation and completing homework.  Preparing these solved questions will help with homework and exams as well.

As mentioned above, Circle is not the most difficult topic that students will have to go through, but it is crucial for students to make sure that they are very clear with the basics. So much so that the concepts and ideas in this chapter must come to them as second nature. 

This chapter forms a base for chapters like Parabola and Ellipse and that is why if students do not have complete clarity in this chapter, then they may have more doubts in these future chapters. Everything is connected when it comes to this chapter and that is why we urge our students to study as much as possible to make sure that they do not miss out on marks from not one but three very important chapters

Importance of RS Aggarwal Solutions Class 11 Circle

• RS Aggarwal Solutions Class 11 Circle helps to prepare for exams without any worries. It will make students work much more manageably.

• With RS Aggarwal’s solutions, a student can learn wisely and work hard.

• Make a routine and allot time for different subjects. Routine makes learning fun and values time. 

• A key to good scores is to revise the previous year’s question paper from RS Aggarwal’s solution.

• RS Aggarwal Solutions Class 11 Circle is portrayed in a much simpler manner. 

• It provides all the short answer-type questions and Solved examples for the student guide.

These points explain how studying from  RS Aggarwal Class 11 Solutions Chapter-21 Circle helps a student in their goal to score the most marks. Therefore, keeping these things in mind, students must approach solving problems from the textbook in the same manner. The problems will not be very simple, but that is why having solutions to them is key. You can use these solutions anywhere anytime.

At Vedantu, we believe in studying in the best environment according to you and that is why we highly recommend printing these solutions on a sheet of paper or carrying the PDF with you on your phone at all times. Download this solution today and compare your answers with that of the experts. Find out where you need to work more and build a strong conceptual platform to solve all the problems related to circles. 

Solved Examples: 

1) When the center of a circle is not at the origin, Find the Equation?

Ans: Let C(h, k) be the center of the circle.

P(x, y) be any point on the circle.

Therefore, the radius of a circle is t.

By using the distance formula,

(x-a)2 + (y-b)2 = t2

Let radius be ‘t.’

Therefore, the equation of the circle with center (h, k). 

The radius ‘t’ is,

(x-a)2+(y-b)2 = t2

which is the standard form for the equation of a circle.

2) What are the Critical Points for Solving this Chapter?

Ans: 

1. The general form of the equation of a circle : (x – a)2 + (y – b)2 = r2

2. The equation of a circle with radius and center at the origin r is given by

x2 + y2 = r2

3. The equation of a circle with A(x1 , y1 ), B(x2 , y2) as the endpoints of a diameter is given by (x – x1)(x – x2) + (y – y1)(y – y2) = 0.

3) Consider a Circle that has a Center at the Origin and a Radius is Equal to 8 Units.

Solution:

Given: Centre is (0, 0), radius is 8 units.

We know that the equation of a circle when the center is origin:

x2+ y2 = t2

For the given condition, the equation of a circle 

x2 + y2 = 82

x2 + y2= 64, which is the equation of a circle

4) Find the Circle Equation that has the Center at (3,5), and the Radius is 4 Units.

Solution:

Here, the center of the circle is not an origin.

Therefore, the general equation of the circle is,

(x-3)2 + (y-5)2 = 42

x2 – 6x + 9 + y2 -10y +25 = 16

x2 +y2 -6x -10y + 18 =0