Circle
Circle holds a high pedestal in 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 a prerequisite to study Circles.
In this chapter, we discuss the algebraic equations representing a circle and the lines associated with it i.e. a tangent, a pair of tangents and a chord of contact. The concepts of family of circles and that of common tangents to two circles under various configurations are given due importance.
A circle can be defined as the curve traced out by a point that moves so that its distance from a given point is constant. The distance between any of the points and the center is called the radius. It can also be defined as the locus of a point equidistant from a fixed point.
The general equation of a circle in Cartesian coordinates with canter at (a, b) and radius r is given by
If the center is at (0, 0) then the equation becomes
In parametric form it may be represented as
where t is a parameter ranging between 0 to 2π.
We first recall the important terms and formulae related to circles:
· Length of circumference in a circle is given by 2πr, where r is the radius.
· The diameter is the longest chord of the circle.
 Area of a circle is given by πr^{2}.
 The circle which is centered at the origin with radius 1 is called the unit circle.
 A line segment whose end points lie on a circle is defined as a chord.
 A line that cuts a circle at two distinct points is called a secant.
 Chords are equidistant from the center of a circle if and only if they are equal in length.
 Equal chords of a circle subtend equal angles at the center
 An arc is a section of the circumference of a circle.
 A sector is an area of a circle bounded by 2 radii.
 A quadrant is onequarter of a circle.
 A segment of a circle is the part bounded by a chord and the circumference.
 Equal arcs of a circle subtend equal angles at the center.
 An angle in a semicircle is a right angle.
Theorem:
An angle at the circumference of a circle is half the angle at the center subtended by the same arc.
Proof:
Let AB be an arc of a circle with center O, and let P be any point on the opposite arc. We need to prove AOB = 2APB. The proof can be carried on in three cases:
Case 1: O lies inside APB
Case 2: O lies on APB
Case 3: O lies outside APB
We shall prove the case 1 here and the rest cases follow on the same lines.
Case 1: Join PO and produce PO to Q. Then OA = OB = OP (radii), so we have two isosceles triangles OAP and OAQ.

Let


OAP

= α and OBP = β.



Then


APO

= α

(base angles of isosceles OAP)


and


BPO

= β

(base angles of isosceles OBP).


Hence


AOQ

= 2α

(exterior angle of OAP)


and


BOQ

= 2β

(exterior angle of OBP),




AOB

= 2α + 2β = 2(α + β) = 2 × APB.

Result: Angle between the tangent and the radius is 90°.
Result: Angles in the same segment are equal.
Result: Angle in a semicircle is 90°.
Result: Two angles at the circumference subtended by the same arc are equal.
You can get more clarity by viewing the video
You may consult the past year papers to get an idea about the type of questions asked.
Illustration 1: The angle between a pair of tangents drawn from a point P to the circle x^{2} + y^{2} + 4x  6y + 9 sin^{2}๐ +13 cos ^{2}๐ = 0 is 2๐. The equation of the locus of the point P is
1. x^{2} + y^{2} + 4x  6y + 4 = 0
2. x^{2} + y^{2} + 4x  6y – 9 = 0
3. x^{2} + y^{2} + 4x  6y  4 = 0
4. x^{2} + y^{2} + 4x  6y + 9 = 0
Solution: The equation of the circle is x^{2} + y^{2} + 4x  6y + 9 sin^{2}๐ +13 cos ^{2}๐ = 0
Hence, the center of the circle = (2, 3)
Radius of the circle is √ (2)^{2} + (3^{2}) – 9 sin^{2}๐ 13cos^{2}๐
= √13 – 9 sin^{2}๐ 13cos^{2}๐
= √13sin^{2}๐ – 9 sin^{2}๐
= √4 sin^{2}๐ = 2 sin ๐
Now, sin ๐ = OA/OP = 2 sin ๐ / √ (h+2)^{2} + (k3)^{2}
Hence, (h+2)^{2} + (k3)^{2} = 4
h^{2} +k^{2} + 4h 6k +9 = 0.
Therefore, the locus of P is x^{2} +y^{2} + 4x  6y +9 = 0.
Illustration 2: Tangents drawn from the point P(1,8) to the circle
x^{2} + y^{2}  6x  4y  11 = 0 touch the circles at the points A and B. The equation of the circumcircle of the triangle PAB is
1. x^{2} + y^{2} + 4x  6y + 19 = 0
2. x^{2} + y^{2}  4x  10y + 19 = 0
3. x^{2} + y^{2}  2x + 6y  29 = 0
4. x^{2} + y^{2}  6x  4y + 19 = 0
Solution: The equation of the circle is x^{2} + y^{2}  6x  4y  11 = 0
Points P, A, C and B are given to be concyclic as angle PAC + angle PBC = π
Hence, the circumcircle of โPAB will pass through C.
As angle PAC = π/2, we can use the diametric form where PC is the diameter of the circumcircle of โPAB.
Using diametric form,
(x1)(x3) + (y8) (y2) = 0.
x^{2} + y^{2}  4x  10y + 19 = 0
Illustration 3: From the origin chords are drawn to the circle (x1)^{2} + y^{2} = 1. The equation of the locus of the midpoints of these chords is…
Solution: The given circle is x^{2} + y^{2} – x = 0.
Let B (h, k) be the midpoint of a chord OA of the circle x^{2} + y^{2} – 2x = 0.
Hence, the equation of the chord bisected at the point is S_{1} =T.
Hence, h^{2} +k^{2} 2h = xh + yk – (x + h) passes through (0, 0).
So, h = h^{2} + k^{2} 2h
h^{2} + k^{2} –h =0.
Hence, the required locus of P is x^{2} + y^{2} –x = 0.
“Circle” is one of the most scoring and hence an important chapter of
Coordinate Geometry in the
Mathematics syllabus of IIT JEE. There are some fixed patterns of questions asked from this topic. You are expected to do all the questions based on this to remain competitive in IIT JEE examination.
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