Circles - Equation of Circle under Different Conditions with Examples

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:

\begin{array}{l} x^{2} + y^{2} = r^{2} \end{array}

Centre = (0, 0) and

Radius = r

2. Equation of a circle in centre radius form:

\begin{array}{l} {(x - h)}^{2} + {(y - k)}^{2} = r^{2} \end{array}

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

3. Equation of a circle in general form:

\begin{array}{l} x^{2} + y^{2} + 2 g x + 2 f y + c = 0 \end{array}

Centre = (–g, –f )

r² = g² + f² – c

\begin{array}{l} Radius = \sqrt{g^{2} + f^{2} - c} \end{array}

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):

\begin{array}{l} | \begin{array}{c} x^{2} + y^{2} & x & y & 1 \\ x_{1}^{2} + y_{1}^{2} & x_{1} & y_{1} & 1 \\ x_{2}^{2} + y_{2}^{2} & x_{2} & y_{2} & 1 \\ x_{3}^{2} + y_{3}^{2} & x_{3} & y_{3} & 1 \end{array} | = 0 \end{array}

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

\begin{array}{l} {(x - a)}^{2} + {(y - a)}^{2} = a^{2} \end{array}

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

Equation of circle touching x axis

\begin{array}{l} {(x - α)}^{2} + {(y - a)}^{2} = a^{2} \end{array}

Touches y-axis only at (a, β)

\begin{array}{l} {(x - a)}^{2} + {(y - β)}^{2} = a^{2} \end{array}

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

\begin{array}{l} x^{2} + y^{2} - α x - β x = 0 \end{array}

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)

Parametric equation of a circle

= 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.

Position of a point with respect to circle

R – radius

cp > R , {point lie outside}

cp = R , {on the curve}

cp < R , {inside the curve}

Parametric Equation of a Circle – Video Lesson

Parametric Equation of a Circle

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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

\begin{array}{l} x x_{1}^{} + y y_{1} + g (x + x_{1}) + f (y + y_{1}) + c = 0 \end{array}

Tangent having slope ‘m’

y = mx + C

Where,

\begin{array}{l} c = \pm (\sqrt{g^{2} + f^{2} - c}) (\sqrt{1 + m^{2}}) \end{array}

\begin{array}{l} c = \pm \sqrt[r]{1 + m^{2}} \end{array}

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

T² = ss₁

Where,

\begin{array}{l} T \equiv x x_{1} + y y_{1} + g (x + x_{1}) + f (y + y_{1}) \end{array}

\begin{array}{l} S = x^{2} + y^{2} + 2 g x + 2 f y + c = 0 \end{array}

and

\begin{array}{l} S_{1} \equiv x {_{1}}^{2} + {y_{1}}^{2} + 2 g x_{1} + 2 f y_{1} + c = 0 \end{array}

Equation of tangent of circle

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

\begin{array}{l} S \equiv x^{2} + y^{2} + 2 g x + 2 f y + c = 0 \end{array}

is

\begin{array}{l} \frac{x - x_{1}}{x_{1} + g} = \frac{y - y_{1}}{y_{1} + f} \end{array}

Equation of Chord

Equation of chord

Equation of chord PQ

\begin{array}{l} w h e r e_{↓} T = S_{1} \end{array}

\begin{array}{l} T \equiv x x_{1} + y y_{1} + g (x + x_{1}) + f (y + y_{1}) + c = 0 \end{array}

\begin{array}{l} S_{1} \equiv x_{1}^{2} + y_{1}^{2} + 2 g x_{1} + 2 f y_{1} + c = 0 \end{array}

Chord of Contact

Chord of contact

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

\begin{array}{l} x x_{1} + y y_{1} + g (x + x_{1}) + f (y + y_{1}) + c = 0 \end{array}

Radical Axis to the Two Circles

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

\begin{array}{l} S_{1} \equiv x^{2} + y^{2} + 2 g_{1} x + 2 f_{1} y + c_{1} = 0 \end{array}

and

\begin{array}{l} S_{2} = x^{2} + y^{2} + 2 g_{2} x + 2 f_{2} y + c_{2} = 0 \end{array}

The equation of the radical axis is

S₁ – S₂ = 0

Family of Circles

S₁ + λS₂ = 0

Where ‘λ’ is the parameter

Family of circles

Particular Cases of Circle

Particular Cases of Circle

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Problems on Circles

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

\begin{array}{l} 3 x^{2} + 3 y^{2} - 8 x - 10 y + 3 = 0. \end{array}

Solution:

We rewrite the given equation as

\begin{array}{l} x^{2} + y^{2} - \frac{8}{3} x - \frac{10}{3} y + 1 = 0 \end{array}

\begin{array}{l} \Rightarrow g = - \frac{4}{3}, f = - \frac{5}{3}, c = 1. \end{array}

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

\begin{array}{l} \sqrt{\frac{16}{9} + \frac{25}{9} - 1} = \sqrt{\frac{32}{9}} = \frac{4 \sqrt{2}}{3} . \end{array}

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

\begin{array}{l} \sqrt{{(4 - 1)}^{2} + {(6 - 2)}^{2}} = \sqrt{25} = 5. \end{array}

Hence, the equation of the circle is

\begin{array}{l} {(x - 1)}^{2} + {(y - 2)}^{2} = 25 \end{array}

\begin{array}{l} \Rightarrow x^{2} + y^{2} - 2 x - 4 y = 20. \end{array}

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

\begin{array}{l} (x + 4) (x - 12) + (y - 3) (y + 1) = 0. \end{array}

On the y-axis,

\begin{array}{l} x = 0 \Rightarrow - 48 + y^{2} - 2 y - 3 = 0 \end{array}

\begin{array}{l} \Rightarrow y^{2} - 2 y - 51 = 0 \Rightarrow y = 1 \pm \sqrt{52} \end{array}

Hence, the length of intercept on the y-axis

\begin{array}{l} = 2 \sqrt{52} = 4 \sqrt{13} . \end{array}

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

Solution:

Let the equation be

\begin{array}{l} x^{2} + y^{2} + 2 g x + 2 f y + c = 0. \end{array}

Substituting the coordinates of the three given points, we get

\begin{array}{l} 2 g + 2 f + c = - 2, \end{array}

\begin{array}{l} 4 g - 2 f + c = - 5, \end{array}

\begin{array}{l} 6 g + 4 f + c = - 13. \end{array}

Solving the above three equations, we obtain

\begin{array}{l} f = - 1 / 2; g = - 5 / 2, c = 4. \end{array}

Hence, the equation of the circle is

\begin{array}{l} x^{2} + y^{2} - 5 x - y + 4 = 0. \end{array}

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

\begin{array}{l} = | \frac{15 + 48 - 1}{\sqrt{25 + 144}} | = \frac{62}{13} . \end{array}

Hence, the required equation of the circle is

\begin{array}{l} {(x - 3)}^{2} + {(y - 4)}^{2} = {(\frac{62}{13})}^{2} \end{array}

\begin{array}{l} \Rightarrow x^{2} + y^{2} - 6 x - 8 y + \frac{381}{169} = 0. \end{array}

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

\begin{array}{l} x^{2} + y^{2} - 4 x - 2 y - 20 = 0. \end{array}

Solution:

Since

\begin{array}{l} S_{1} = 10^{2} + 7^{2} - 4 \times 10 - 2 \times 7 - 20 > 0, \end{array}

P lies outside the circle.

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,

\begin{array}{l} P C = \sqrt{{(10 - 2)}^{2} + {(7 - 1)}^{2}} = 10 \end{array}

and CB = radius

\begin{array}{l} = \sqrt{4 + 1 + 20} = 5 \end{array}

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

\begin{array}{l} y - 3 = \frac{3 - 1}{4 - 2} (x - 4) \end{array}

, i.e., x – y – 1 = 0.

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

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

\begin{array}{l} = \sqrt{2} \end{array}

Hence, the equation of circle is

\begin{array}{l} {(x - 1)}^{2} + {(y - 0)}^{2} = 2 \end{array}

or

\begin{array}{l} x^{2} + y^{2} - 2 x - 1 = 0. \end{array}

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

\begin{array}{l} x^{2} + y^{2} + 2 x + 6 y = 0 \end{array}

and

\begin{array}{l} x^{2} + y^{2} - 4 x - 2 y - 6 = 0 \end{array}

Solution:

Equation of common chord is

\begin{array}{l} S_{1} - S_{2} = 0 \end{array}

or

\begin{array}{l} 6 x + 8 y + 6 = 0 \end{array}

or

\begin{array}{l} 3 x + 4 y + 3 = 0. \end{array}

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

\begin{array}{l} \sqrt{1 + 9} = \sqrt{10} . \end{array}

Distance of C₁ from the common chord

\begin{array}{l} = \frac{| - 3 - 12 + 3 |}{\sqrt{9 + 16}} = \frac{12}{5} \end{array}

Therefore, the length of common chord

\begin{array}{l} = 2 \sqrt{{(\sqrt{10})}^{2} - {(\frac{12}{5})}^{2}} = 2 \sqrt{10 - \frac{144}{25}} = 2 \sqrt{\frac{106}{25}} = \frac{2 \sqrt{106}}{5} \end{array}

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

\begin{array}{l} x^{2} + y^{2} = 4 \end{array}

homogeneous with respect to line

\begin{array}{l} x + y = a, \end{array}

we should have

\begin{array}{l} x^{2} + y^{2} - 4 {(\frac{x + y}{a})}^{2} = 0 \end{array}

\begin{array}{l} \Rightarrow a^{2} (x^{2} + y^{2}) - 4 (x^{2} + y^{2} + 2 x y) = 0 \end{array}

\begin{array}{l} \Rightarrow x^{2} (a^{2} - 4) + y^{2} (a^{2} - 4) - 8 x y = 0 \end{array}

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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Frequently Asked Questions

Q1

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 x² + y² = r².

Q2

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.

Q3

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 = r².

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