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 )

r^{2} = g^{2} + f^{2} – c

**4. Equation of a circle with points P(x _{1}, y_{1}) and Q(x_{2}, y_{2}) as extremities of diameter:**

(x – x_{1}) (x – x_{2}) + (y – y_{1}) (y – y_{2}) = 0

**5. Equation of circle through three non-collinear points** P(x_{1}, y_{1}), Q(x_{2}, y_{2}) and R(x_{3}, y_{3}):

Area of circle = πr^{2}

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 = x^{2} + y^{2} = r^{2}

X = r cos θ

Y = r sin θ

Squaring both sides,

x^{2} + y^{2} = (r^{2} cos^{2}θ + r^{2} sin^{2}θ)

= r^{2} (cos^{2}θ + sin^{2}θ)

x^{2} + y^{2} = r^{2}

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

Let the circle be x^{2} + y^{2} + 2gx + 2fy + c = 0 and p(x_{1}, y_{1}) 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

x^{2} + y^{2} + 2gx + 2fy + c = 0

A tangent at point P(x_{1}, y_{1}).

### Equation of Tangent of Circle

Tangent having slope ‘m’

y = mx + C

Where,

Pair of tangents from external point p (x_{1}, y_{1})

T^{2} = ss_{1}

Where,

Equation of normal at p(x_{1}, y_{1}) 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 S_{1} and S_{2,}

The equation of the radical axis is

S_{1} – S_{2} = 0

### Family of Circles

S_{1} + λS_{2} = 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 S_{1} is C_{1}(-1, -3) and its radius is

Distance of C_{1} 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

∴ a^{2} – 4 + a^{2} – 4 = 0

⇒ a^{2} – 4 = 0

⇒ a = ±2

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

Hence, the correct answer is (a).

## Circles – Important Topics

## Circles – Class 11 Maths Chapter 11 Quiz

### Circles – Important Questions

### Circles – Solved Problems

### Straight Lines and Circles – Top 12 Important Questions

### Straight Lines & Circles JEE Advanced Questions

### Circles – One Shot Revision

## Frequently Asked Questions

### Give the standard equation of a circle with centre (0, 0) and radius r.

The standard equation of a circle with centre (0, 0) and radius r is given by x^{2} + y^{2} = r^{2}.

### What do you mean by a chord of a circle?

A chord is a line segment joining two points on a circle. The diameter is the largest chord.

### Give the standard equation of a circle with centre (h, k) and radius r.

The standard equation of a circle with centre (h, k) and radius r is given by (x-h)^{2} + (y-k)^{2} = r^{2}.

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