The post JEE Conic Sections: Ellipse appeared first on Magoosh JEE Blog.
]]>So, let’s learn about ellipses in this blog post!
Hey there! Let’s start this chapter with a small exercise. No, not the JEE problems, but a simple, fun exercise. Pick up any coin from your wallet. Found it? Now switch off the lights and shine a torch or flashlight onto the coin pressed against a wall. Slowly pull the coin away from the wall. You’d see a shadow of the circle. Now, rotate the coin slightly in any direction. What shaped shadow did you observe? It looks like a squashed circle, doesn’t it? Well, we have a word for it–ellipse!
Ellipse is a part of conic sections, that is, it can be obtained as a cross-section of a cone. If you cut a cone parallel to its base, you get a circle. But if you make a cut on the cone at an angle through its curved face, you get an ellipse, as shown in the figure below. More precisely, an ellipse is any plane section of a cone, not containing the cone’s apex, and with a slope less than the slope of the lines on the cone.
In fact, you could call a circle a special type of ellipse!
Now that we have a brief idea of what an ellipse is and what it looks like, let’s describe the ellipse mathematically.
Ellipse can be described as a curve in which the ratio of the distance between any point P on the curve and a fixed point F, and the distance between that point (P) and a fixed line L is a constant real number greater than/equal to zero and less than one.
Too much jargon? Don’t worry. Let’s break it down step-by-step.
Take a fixed point F, say (0, 0).
Now take a fixed line L, say x = 1, or, x — 1 = 0.
Now, take a general point P(x, y). Drop a perpendicular from point P onto line L. Let the perpendicular from P and line L intersect at P’, that is, the (perpendicular) distance between P and L is PP’. The distance between P and F is PF.
We have,
PF/PP’ = constant ∊ (0, 1). Let’s set this constant to be 0.5.
⇒PF/PP’ = 0.5
⇒√((x — 0)^{2} + (y — 0)^{2})/(|x — 1|/√(1^{2})) = 0.5
⇒2√(x^{2} + y^{2}) = |x — 1|
⇒4(x^{2} + y^{2}) = (x — 1)^{2}
⇒4x^{2} + 4y^{2} = x^{2}–2x + 1
⇒3x^{2} + 4y^{2} + 2x — 1 = 0 is the equation of the ellipse.
If you sketch the curve, you’d get something like this:
The curve in green is the desired ellipse. Every point on this curve is twice as far from the line x — 1 = 0 as it is from the point (0, 0). You can check and verify it for any point on this ellipse.
The fixed point F is known as ‘focus’ of the ellipse, and the fixed line L is known as the ‘directrix’ of the ellipse.
The ratio, (Distance of any point from Focus)/(Distance of the same point from Directrix) is known as the eccentricity of the ellipse, denoted by ‘e’, where 0 < = e < 1 Eccentricity is a measure of the degree of resemblance of an ellipse to a circle. For a circle, eccentricity will be zero. That is why we consider a circle as a special type of ellipse.
Every ellipse has 2 foci and 2 corresponding directrices, due to symmetry. Given a focus and a directrix, we can uniquely define an ellipse. Likewise, an ellipse can be uniquely defined by giving the focus and its eccentricity.
Ellipse can also be described in another, simpler way. Ellipse can be described as a curve in which the sum of distances between any point P on the curve and 2 fixed points is a constant number, greater than the distance between the fixed points.
Got confused? No worries, let’s simplify this as well.
Take 2 points, say F1(–4, 0) and F2(4, 0). Now, take a general point P(x, y). The distance of the point P from F1 is given by PF1 = √((x– (–4))^{2} +(y — 0)^{2}). Similarly, the distance of the point P from F2 is given by PF2 = √((x — 4)^{2} + (y — 0)^{2}). The distance between F1 and F2 is √((4– (–4))^{2} + (0 — 0)^{2}) = 8 units.
We have, PF1 + PF2 = constant, where constant > F1F2
⇒√((x– (–4))^{2} + (y-0)^{2}) + √((x — 4)^{2} + (y — 0)^{2}) = constant, constant > 8
Let’s take constant = 10
⇒√(x^{2} + 8x + 16 + y^{2}) + √(x^{2} –8x + 16 + y^{2}) = 10
⇒√(x^{2} + 8x + 16 + y^{2}) = 10 — √(x^{2} — 8x + 16 + y^{2})
⇒x^{2} + 8x + 16 + y^{2} = 100 + x^{2} — 8x + 16 + y^{2} — 20√(x^{2} –8x + 16 + y^{2})
⇒25 — 4x = 5√(x^{2} –8x + 16 + y^{2})
⇒625 + 16x^{2} — 200x = 25x^{2} — 200x + 400 + 25y^{2}
⇒9x^{2} + 25y^{2} — 225 = 0 is the desired ellipse
If you take any point P(x, y) located on this ellipse, the sum of its distances from the points F1(–4, 0) and F2(4, 0) will be 10. Try to check and confirm it!
In general, for any ellipse, the fixed points F1 and F2 are known as the ‘foci’ of the ellipse. Yes, these are the same foci that we discussed in the previous description of foci and directrices. Isn’t it interesting? Hold your excitement, there is a lot more interesting stuff coming up.
Also, the fixed sum of distances is known as the length of ‘major axis’ of the ellipse, denoted by ‘2a’. The major axis is the line joining the two farthest points in a given ellipse. For example, in the ellipse described above, the x-axis is the major axis as it is the line joining (–5, 0) and (5, 0), which are clearly the farthest points in that ellipse. Length of the major axis is the distance between the farthest points, given by √((5–(–5))^{2} + (0 — 0)^{2}) = 10 units.
Also, the ratio (F1F2)/(2a) is nothing but the eccentricity (e) of the given ellipse, that is,
e = (F1F2)/(2a) = 8/10 = 0.8
Note that this is the same eccentricity about which we learned in the previous section. It comes full circle.
Let’s quickly summarize our discussion on ellipse,
In this post, we learned some important theoretical concepts of an Ellipse. Hope you found it useful. For more amazing JEE study material, check out Magoosh JEE Product. Happy learning!
For more information about JEE Conic Sections, check out the following Magoosh resources:
Additional resources:
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]]>The mathematical concept of Hyperbola is essential for engineering and architectural studies. It’s quite obvious that Hyperbola is very important for JEE. All its properties must be at your fingertips in order to solve the questions asked in JEE.
You can think of a hyperbola to be two curves that resemble infinite bows.
A hyperbola is the locus of all points in a plane, the difference of whose distances from two fixed points (focus) is constant.
Now let me introduce you to the various terms related to hyperbola:
Ax^{2} + Bxy + Cy^{2} + Dx + Ey + F = 0
The expression B^{2} — 4AC is called the discriminant, which is used to determine the type of conic section.
For a hyperbola, B^{2} — 4AC > 0.
A hyperbola with Center as Origin (0, 0) and both the foci on the x-axis is given by the equation:
The properties of the above hyperbola are:
Now you might be wondering about ‘e’ used in various formulae. ‘e’ is known as the Eccentricity of a conic section. It is the ratio between the focus and the directrix which remains constant for a given conic section. For a hyperbola, e > 1.
As the name suggests, it is the distance of any arbitrary point on the hyperbola from the focus.
Let P be any point present on the hyperbola,
|S_{1}P — S_{2}P| = 2a
Clearly, the focal distance of any point on the hyperbola is constant and is equal to the length of the Transverse Axis.
The directions of the axes should be parallel to the co-ordinate axis.
The parametric equations of the general hyperbola being discussed are:
x = a sec θ , y = b tan θ
Or
x = a cosh θ , y = b sinh θ
1) The equation of a chord joining two points P( a sec θ₁ , b tan θ₁) and Q( a sec θ₂ , b tan θ₂) on the hyperbola
2) The equation of the chord of the hyperbola bisected at point (x₁, y₁) is given by,
T = S₁
3) Equations of the chord of contact of tangents drawn from a point (x₁, y₁) to the hyperbola is given by,
1) In point form the equation of the tangent to the hyperbola is,
2) In parametric form the equation of the tangent to the hyperbola is,
3) In Slope form the equations of the tangent to the hyperbola is,
y = mx ± √(a²m² — b²)
Here, m is the slope of the normal the tangent to the hyperbola.
4) The line y = mx + c touches the hyperbola if c² = a²m² — b².
1) Point Form: In point form the equation of normal to the hyperbola is,
2) Parametric Form: The equation of normal to the hyperbola at a point P(a secθ, b tanθ) is,
ax cosθ + by cotθ = a² + b²
3) Slope Form: The equation of normal to the hyperbola is,
Here, m is the slope of the normal to the hyperbola being discussed.
4) A maximum of 4 normals can be drawn to a hyperbola from a point P(x₁, y₁)
1) Diameter: Diameter is the locus of mid-points of all the parallel chords of the hyperbola. The equation of the diameter, bisecting a system of parallel chords with each of slope m to the hyperbola is given by,
2) Conjugate Diameter: The diameters of a hyperbola are said to be conjugate if each diameter bisects the chords parallel to the other.
The diameters y = m₁x and y = m₂x are conjugate if,
m₁m₂ = b²/a²
3) In a pair of conjugate diameters of a hyperbola, only one of them meets the hyperbola in real points while the other meets the hyperbola at imaginary points.
The locus of points of intersection of the tangents to the hyperbola, which are perpendicular to each other.
The equation of Director Circle is,
x² + y² = a² — b²
A hyperbola for which the asymptotes are perpendicular is known as a rectangular hyperbola.
In other words, it is a hyperbola with a = b.
Therefore, the eccentricity of a rectangular hyperbola is √2 .
The equation for a rectangular hyperbola can be reduced to,
xy = a²
1) Asymptotes are the co-ordinate axes, x = 0 and y = 0.
2) e = √2.
3) Center is O(0, 0).
4) Foci are S(√2c,√2c) and S₁(–√2c, –√2c).
5) Vertices are A(c, c) and A₁(–c, –c).
6) Length of Latus Rectum = 2√2c.
7) In parametric form, the equation is x = ct and y = c/t.
2) The combined equation of pair of tangents drawn from an external point P(x₁,y₁)
is SS₁–T².
3) The equation of a chord of the hyperbola whose mid-point is (x₁,y₁) is given by
T = S₁.
4) Eccentricity of a Rectangular Hyperbola is √2 and the angle between asymptotes is 90°.
5) If a triangle is inscribed in a hyperbola, then it’s orthocenter lies on the hyperbola.
In this blog post, you learnt about some important concepts about Hyperbola. Hope you found it useful. For more amazing JEE study material, check out Magoosh JEE Product. Happy learning!
For more information about JEE Conic Sections, check out the following Magoosh resources:
Additional resources:
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]]>Let’s start by picturing two lines intersecting each other. Now make one of the lines (called the Generating Line) revolve around the other. We get what is called a double cone. Let another plane now intersect it (obviously, not where the cone’s vertex lies–in which case you get straight lines!!).
What you get on the plane is called a conic section. The following are five conic sections and they are characterized by what is known as eccentricity (often denoted by ‘e’).
So let’s get started!
We are given a straight line (directrix) and a point (focus). Then on a plane, the parabola is the set of all points such that they are equidistant from the directrix and the focus. The vertex is the point closest to the directrix. A line parallel to the directrix and passing through the focus cuts a conic section in two points, right? This particular line segment is called the latus rectum.
Taking the directrix as x + a = 0 and the focus as (a, 0), the equation of a right open horizontal parabola is: y^{2} = 4 a x
Thus, the parametric coordinates are: (at^{2}, 2at)
For latus rectum, we set x = a and get: l = 4 a
It can be easily shown that a point (x_{1}, y_{1}) is an interior point if S_{1} > 0, lies on the parabola if S_{1} = 0 and exterior to it if S _{1} < 0.
Focal Chord:
Any chord of the focus of a parabola is called a focal chord. It can be easily proved that if t_{1} and t_{2} are the parameters for its endpoints, then t_{1} t_{2} = –1.
(Complete description using Desmos Graphing Calculator)
The point of intersection of two tangents (drawn at t_{1} and t_{2}) is (at_{1}t_{2}, a(t_{1}+t_{2})).
Now, I am about to tell you something interesting! Just try to think over it. The tangents that are drawn at the end of a focal chord intersect at right angles on the directrix (Hint: t_{1}t_{2} = –1 and hence m_{1}m_{2} = –1). This also means that a circle drawn with a focal chord as the diameter touches the directrix! Don’t worry, see a complete proof here.
For any exterior point (x_{1}, y_{1}), the equation T^{2} = SS_{1} gives the tangents to the parabola from the given point. When S_{1} = 0, the point lies on the parabola and T = 0 which is actually true. Also notice that if S_{1} < 0, then T^{2} >0, and hence no tangent exists from the interior of a parabola.
When the tangents from an exterior point (x_{1}, y_{1}) are drawn, the points where they touch can be joined to form ‘The Chord of Contact’: T_{1} = 0
When an interior point (x_{1},y_{1}) is given, then the chord through the point such it is the midpoint is given by:
T = S_{1}
Don’t worry. The following examples will clear all your doubts.
Question 1
For the parabola y^{2} = 16 x, determine the angle between the tangents drawn from (–8,4).
Answer: c) 72°
Solution:
For the parabola y^{2} = 4 (4) x, a = 4
Putting (x_{1}, y_{1}) = (–8, 4) in the equation T^{2} = SS_{1} gives:
(4y — 2(4)(x — 8))^{2} = (y^{2} — 16x)(4^{2} — 16(–8))
=> x^{2} — xy –2y^{2} + 20x + 8y + 64 = 0
=> (x + y + 4)(x — 2y + 16) = 0
Hence, the slopes are m_{1} = ½ and m_{2} = –1
The angle between the tangents is tanθ = mod((m_{2} — m_{1})/(1 — m_{1}m_{2})) = 3
Since tanθ > 1 => θ > 45o
Also, θ < 90^{o}
Checking the options, the option (c) is the most appropriate.
You should try more and more problems on this topic. After all, practice makes a man perfect!
Moving on, let us study the normal to a parabola!
(Visualizing normals using Desmos Graphing Calculator)
Thus, if the slope of a normal is given, the equation of the normal can be rewritten as:
y = m x — (a m^{3} + 2 a m)
Moreover, you can even find the point to which the normal corresponds if its slope is given (using the fact that t = –m).
(at^{2}, 2at) = (am^{2}, –2am)
If you are given a point (x_{1}, y_{1}), then you can also find the possible slopes for a normal at that point.
y = m x — (am^{3} + 2am) ⇒ am^{3} + m (2a — x) + y = 0
Interesting! Through a single point on a parabola, as many as three normal could pass!
Reflection property of a parabola: A ray of light parallel to the axis of parabola reflects and passes through the focus. This reflection property is exploited to make spherical mirrors used in physics to remove any spherical aberrations.
In this blog post, you learnt about some important concepts about Parabola. Hope you found it useful. For more amazing JEE study material, check out Magoosh JEE Product. Happy learning!
For more information about JEE Conic Sections, check out the following Magoosh resources:
Additional resources:
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]]>Coordinate geometry carries huge weightage in the JEE mathematics syllabus, and conic sections is an important topic in JEE coordinate geometry. You can expect around two to five questions on this topic in the JEE paper. So, it is very important for you to have good command over this topic if you want to score high. Read on to know some important things about the JEE conic sections!
Technically, you would define a conic section as a curve formed by the intersection of a plane with the surface of a double-napped right circular cone. And if you don’t know what a double-napped right circular cone is—let’s make it simple.
Take a right circular cone and stack an inverted right circular cone on its vertex. What you get is a double-napped right circular cone. When you take into account all possible orientations of the intersecting plane with the double-napped cone, you get four curves, namely: a circle, an ellipse, a parabola, and a hyperbola.
There is another definition that you will find more useful for the JEE. A conic section is a locus of all points P such that its distance from a fixed point F (called the focus of the conic), is a constant multiple e, the eccentricity, of the distance from P to a fixed line L, called the directrix of the conic.
This gives us the working mathematical form for a conic section,
Now, depending on the value of e, the curves exhibit varying properties. If 0<e<1, we get a closed curve called ellipse. When e=1, we get a parabola and for e >1, we have a hyperbola.
Question 1
Here is a question that will make you more comfortable with the concept.
Given the equation Identify if it is a
A) Parabola
B) Hyperbola
C) Circle
D) Ellipse
The correct answer is A). It is parabola. With what I told you above, you may find the question mind-boggling! If not, you are really good in algebraic manipulation. Anyway, let me show you how to do it. The aim is to convert the equation 16x^{2} + 9y^{2} –92x –256y –24xy + 76 = 0 into generic conic section formula . We proceed as follows.
You can compare it with the general form. What you get is focus F ≡(1,2) and the equation of directrix 3x + 4y + 7. Also, the eccentricity is 1. So, you can safely say that it is a parabola.
At times, the necessary factorization may not be trivial. In fact, you may not find one at all. In such a case, we employ a different method. The generic equation for a conic section is a general second-degree equation Ax^{2} + 2Bxy + Cy^{2} + 2Dx + 2Ey + F = 0.
In the light of the above conditions, you can easily determine what conic section the equation
16x^{2} + 9y^{2} –92x –256y –24xy + 76 = 0 represents.
B^{2} — 4AC = 24^{2} — 4 x 16 x 9 = 576 — 576 = 0. Hence, a parabola.
Now, looking at the general second-order equation, does it remind you of something else you may have studied earlier? To make things more interesting, it is also the equation for a pair of straight lines. You may be wondering how a single equation could represent both the pair of straight lines and conic sections. Well, give the statement some thought.
A pair of straight lines is itself a conic section. Confused? Don’t be. Suppose you have a double-napped right circular cone. Take a plane such that it passes through the vertex of the double-napped cone and also contains its axis. What curve will you get? A pair of straight lines!
For the time being and also for the rest of the article, let us assume that a pair of straight lines and conic sections are two independent entities. Now that you know all this, you want to distinguish a conic section second-order equation from a pair of straight lines second-order equation.
Given an equation ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0, it will represent a pair of straight lines if abc + 2fgh — af^{2} — ch^{2} — bg^{2} = 0, otherwise any of the four conic sections.
Question 2
Here is another example for you to practice.
What does the given equation 5x^{2} — y^{2} –4xy + 8x — 2y +3 = 0 represent?
A) A pair of straight lines
B) Circle
C) Ellipse
D) Parabola
The correct answer is A) a pair of straight lines. You can use the above formulae to identify the nature of the equation. Also, if you factorize it, you will get (5x + y + 3)(x — y + 1) = 0, which are the equations of the two lines the equation represents.
So far, we have only talked about the classification of conic sections. Sometimes, we are also interested in the coordinates of the center of the conic sections. You would ask why? Well, most of the time the properties we are interested in are unaffected by the transformation of coordinates. In such cases, you can shift the center of the conic to a more desirable coordinate to make calculations simple and easy to perform.
Now that you know how you can exploit it, you need to know how to determine the center of a conic. The easiest way to do so is to partially differentiate the conic equation with respect to x and y, and simultaneously solve for x and y from the two equations obtained.
Consider the general equation ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = f(x,y). Therefore, . When you solve for x and y, you get the coordinate of the center of the conic.
When working with conic sections, there are a few terms that you should know.
Principal Axis: In the case of hyperbola and ellipse, there exist two foci—in contrast to the single focus in a parabola. We define the Principal Axis as a straight line perpendicular to the directrix, which passes through the focus (foci).
Focal Parameter: It is the distance between the focus and the directrix.
Focal Length: For a parabola, the focal length is the distance between the focus and the vertex. For ellipse and hyperbola, it is defined as the distance from the center to any of the two foci.
Latus Rectum: The latus rectum of a conic section is the chord (line segment) that passes through the focus, is perpendicular to the major axis, and has both endpoints on the curve. The latus rectum satisfies some properties which are given below:
You may find it difficult to wrap your head around the generalized form. Studying their properties will give you a hard time if you are allergic to mathematics. In such cases, the standard form of conics come to the rescue.
In all the above equations, the center is (0,0) and the major axis is along x-axis while the minor axis is along y-axis.
In this blog post, we learned about the conic sections, which are present in the JEE maths syllabus. Hope you found it useful. For more amazing JEE study material, check out Magoosh JEE Product. Happy learning!
For more information about JEE Conic Sections, check out the following Magoosh resources:
Additional resources:
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]]>Straight Lines JEE is the foundation of Coordinate Geometry. Coordinate Geometry is that part of JEE Mathematics which almost covers 20% weightage of JEE Advanced Mathematics Paper. This tells us how important Straight Lines JEE is.
Some previously learned concepts that we must know:
To plot any curve, we should have the basic knowledge about the locus of a point, i.e. what constraints the point is being put under, what conditions it should follow. A straight line is the locus of a point which is collinear with two given points.
To study about a straight line you should be well acquainted with terms like:
Note: Slope of vertical line(parallel to y-axis) is not defined, whereas slope of a horizontal line(parallel to x-axis) is zero.
Point to be remembered: Length of an intercept cut is always non-negative whereas Intercept cut by a line can have any Real value. Suppose if in a question you are given that a line has a slope of 1 and has “length of intercept” equals to 1, then there will be two such lines(y=x+1,y=x-1). But if it is said that intercept of the line is 1, then there will be only one such line i.e (y=x+1).
This was all about general terms related to Straight Lines JEE. Now we will be covering sub-topics of Straight Lines JEE which are important.
We can say that the point A(x1,y1) is above the line ax+by+c=0 if
We conclude that the point A(x1,y1) is below the line ax+by+c=0 if
Note: The point A(x1,y1) will be on the line ax+by+c=0 if ax1+by1+c=0.
You should remember this formula otherwise the process is sometimes too lengthy.
There is no need to cram formula for this. We use two basic concepts for this problem.
In this case one can say that image itself lies on the given line. So that point satisfies the equation of given line.
Another equation can be derived from condition of slope.
Of course, there are two angle bisectors given by the equation:
How can we differentiate between the acute angle bisector and the obtuse angle bisector?
It is very simple to determine whether the bisector is an acute angle bisector or an obtuse angle bisector. Compute the angle between the initial line and one of the bisectors. Suppose that angle is θ. Then find the tangent of this angle θ.
In case |tan θ| < 1, then we have 2θ < 90, then this represents the obtuse angle bisector.
On the other hand, if |tan θ| > 1, then this represents the obtuse angle bisector.
It is highly recommended that you should learn the direct results written below:
This concept is used in many problems directly. Rotation of line is also helpful in Optics.
Rotation by θ radians counterclockwise, around the origin gives the new coordinates as:
x=ucosθ+vsinθ
y=−usinθ+vcosθ
These are the important points, tricks which you should keep in mind while solving Straight Lines JEE.
The key to score high in Coordinate Geometry is Practice, Practice and Practice, along with a number of direct results you can remember. You should focus on these points.
The textbook that you can refer for Straight Lines JEE is S.K Goyal. It contains a nice set of questions. One should always attempt previous years’ questions of JEE Main and JEE Advanced. If you are able to solve them by yourself, it will boost your confidence a lot.
Practice hard and we are sure that you will come up with flying colors. There is no shortcut to success. Remember — hard work beats talent, EVERY SINGLE TIME. The more questions you solve, the better you will get at solving questions — which has a direct effect on your JEE rank.
We hope this article helps you.
Cheers!
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]]>JEE Circles is a base to the entire syllabus of Coordinate Geometry in Mathematics. It is the easiest topic of coordinate geometry and with a bit of hard work, it becomes very easy to answer all the questions of this topic.
The preliminary knowledge of the concept of Straight Lines is required to study Circles. Circles introduce a lot of concepts in coordinate geometry. This is also important for complex numbers. Do this topic well. Get good command over locus problems. Practice lots of problems on various forms of equations of a circle. Give special attention to the parametric form.
Study the JEE circles with a good understanding of the diagrams and graphs. This will reduce the effort that is needed to solve any problem in this topic. This is so because having a clear picture of the problem statement in your mind is always helpful.
Determining your approach is as crucial as the process of actually studying. Keep twisting each question you solve. Make room for new possibilities and it will help you to understand it more deeply.
You should first go through the basic concepts through the NCERT/R.D. Sharma and for practicing advanced level, you can try Coordinate Geometry by S.K Goyal (Arihant Publications) or S.L. Loney (Coordinate Geometry). The former is a more JEE oriented and contains much relevant stuff. This is the best book for JEE Circle/Coordinate Geometry, which is written in such a way that even a beginner can understand clearly. It has questions and examples from basic level to advanced level. The solutions to examples are presented in such a manner that you can easily understand. After reading concepts, shortcuts, and examples anyone can solve all the problems with little efforts.
These three are the extremely important properties of circle and questions are asked frequently on these concepts.
With these points in mind, you can score higher in the JEE circles questions. Prepare accordingly and do your best. Good luck!
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