Circles

A circle is the locus of points which moves in a plane such that its distance from a fixed point is always constant. The fixed point is called the ‘centre’ while the fixed distance is called the ‘radius’. This article will help you to have a clear idea of the topics such as circle, equation of tangent, normal and chord of contact. 

Equation of Circle

1. Standard equation of circle:

Centre = (0, 0) and

Radius = r

2. Equation of a circle in centre radius form:

Centre = (h, k), Radius = r.

3. Equation of a circle in general form:

Centre = (–g, –f )

r² = g² + f² – c

4. Equation of a circle with points P(x₁, y₁) and Q(x₂, y₂) as extremities of diameter:

(x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0

5. Equation of circle through three non-collinear points P(x₁, y₁), Q(x₂, y₂) and R(x₃, y₃):

Area of circle = πr²

Perimeter = 2πr, where r is the radius.

Equation of Circle under Different Conditions

• Touches both axes with centre (a, a) and radius r = a

Touches x-axis only, with centre (α, a)

• Touches y-axis only at (a, β)

• Passes through the origin with centre (α/2, β/2)

Parametric Equation of Circle

Equation of circle = x² + y² = r²

X = r cos θ

Y = r sin θ

Squaring both sides,

x² + y² = (r² cos²θ + r² sin²θ)

= r² (cos²θ + sin²θ)

x² + y² = r²

Position of a point w.r.t. to circle

Let the circle be x² + y² + 2gx + 2fy + c = 0 and p(x₁, y₁) be the point.

R – radius

cp > R , {point lie outside}

cp = R , {on the curve}

cp < R , {inside the curve}

Parametric Equation of a Circle – Video Lesson

Equation of Tangents and Normal

The Equation of Tangents and Normal are explained below. Let the equation of the circle be

x² + y² + 2gx + 2fy + c = 0

A tangent at point P(x₁, y₁).

Equation of Tangent of Circle

Tangent having slope ‘m’

y = mx + C

Where,

Pair of tangents from external point p (x₁, y₁)

T² = ss₁

Where,

Equation of normal at p(x₁, y₁) to the circle

Equation of Chord

Equation of chord PQ

Chord of Contact

AB is called the chord of contact. The equation of contact is T = 0.

Radical Axis to the Two Circles

Equation of radical axis to the two circles S₁ and S_2,

The equation of the radical axis is

S₁ – S₂ = 0

Family of Circles

S₁ + λS₂ = 0

Where ‘λ’ is the parameter

Particular Cases of Circle

Problems on Circles

Illustration 1: Find the centre and the radius of the circle

Solution:

We rewrite the given equation as

Hence, the centre is (4/3, 5/3), and the radius (r) is

Illustration 2: Find the equation of the circle with centre (1, 2) and which passes through the point (4, 6).

Solution:

The radius of the circle is

Hence, the equation of the circle is

Illustration 3: Find the equation of the circle whose diameter is the line joining the points (-4, 3) and (12, -1). Find also the length of the intercept made by it on the y-axis.

Solution:

The required equation of the circle is

On the y-axis,

Hence, the length of intercept on the y-axis

Illustration 4: Find the equation of the circle passing through (1, 1), (2, -1) and (3, 2).

Solution:

Let the equation be

Substituting the coordinates of the three given points, we get

Solving the above three equations, we obtain

Hence, the equation of the circle is

Illustration 5: Find the equation of the circle whose centre is (3, 4) and which touches the line 5x + 12y = 1.

Solution:

Let r be the radius of the circle.

Then,

r = distance of the centre, i.e., point (3, 4) from the line 5x + 12y = 1

Hence, the required equation of the circle is

Illustration 6: Find the greatest distance of the point P(10, 7) from the circle

Solution:

Since

Join P with the centre C(2, 1) of the given circle.

Suppose PC cuts the circle at A and B, where A is nearer to C.

Then, PB is the greatest distance of P from the circle.

We have, 

and CB = radius

Therefore, PB = PC + CB = 10 + 5 = 15.

Illustration 7: A foot of the normal from the point (4, 3) to a circle is (2, 1), and the diameter of the circle has the equation 2x – y = 2. Then, find the equation of the circle.

Solution:

The line joining (4, 3) and (2, 1) is also along a diameter.

So, the centre is the point of intersection of the diameter 2x – y = 2 and the line

Solving these, we get the centre as (1, 0).

Also, the radius = the distance between (1, 0) and (2, 1)

Hence, the equation of circle is

Illustration 8: Find the length of the common chord of the circles

Solution:

Equation of common chord is

Centre of S₁ is C₁(-1, -3) and its radius is

Distance of C₁ from the common chord

Therefore, the length of common chord

Illustration 9: A pair of perpendicular straight lines pass through the origin and also through the point of intersection of the curve. The set containing the value of a is

(a) {-2, 2}

(b) {-3, 3}

(c) {-4, 4}

(d) {-5, 5}

Solution:

To make the curve

Since this is a pair of perpendicular straight lines, we have

∴ a² – 4 + a² – 4 = 0

⇒ a² – 4 = 0

⇒ a = ±2

Therefore, the required set of a is {-2, 2}.

Hence, the correct answer is (a).

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