# Equation of circle , Equation of circle under different conditions , Solved Examples

DEFINITION :  A circle is the locus of a point which moves in such a way that its distance from a fixed point , called the center , is always a constant . The distance r from the center is called the radius of the circle .

Twice the radius is known as the diameter d = 2r

The perimeter C of a circle is called the circumference, and is given by

C = πd = 2πr.

The angle a circle subtends from its centre is a full angle equal to 3600 or 2π radians.

### Equation of a circle in various forms :

* The simplest equation of the circle is x2 + y2 = r2 whose centre is (0, 0) and radius r.

* The equation (x − a)2 + (y − b)2 = r2 represents a circle with centre (a, b) and radius r.

* The equation x2 + y2 + 2g x + 2f y + c = 0 is the general equation of a circle with centre (−g , −f) and radius $\displaystyle r = \sqrt{g^2 + f^2 – c}$ .

Case I: If g2 + f2 − c > 0, then real circle is possible.

Case II: If g2 + f2 − c = 0, then the circle formed is called a point circle.

Case III: If g2 + f2 − c < 0, then no real circle is possible.

* Equation of the circle with points P(x1, y1) and Q(x2, y2) as extremities of a diameter is

(x − x1)(x − x2) + (y − y1)(y − y2) = 0.

* The equation of the circle through three non-collinear points P(x1, y1), Q(x2, y2) and R(x3, y3) is

$\large \left| \begin{array}{cccc} x^2 + y^2 & x & y & 1 \\ x_1^2 + y_1^2 & x_1 & y_1 & 1 \\ x_2^2 + y_2^2 & x_2 & y_2 & 1 \\ x_3^2 + y_3^2 & x_3 & y_3 & 1 \end{array} \right| = 0$

Notes:

The general equation of second degree ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents a circle, if

* Coefficient of x2 = coefficient of y2 i.e. a = b

* Coefficient of xy = zero i.e. h = 0.

Equation of a circle under different conditions

### Equation of a circle under different conditions:

(i) Touches both the axes with center (a , a) and radius | a |

(x−a)2 + (y−a)2 = a2

(ii) Touches x-axis only with center (α, a) and radius |a|

(x − α)2 + (y−a)2 = a2

(iii)Touches y–axis only with center (a, β) and radius |a|

(x − a)2 + (y− β)2 = a2

(iv) Passes through the origin with centre (α/2 ,β/2 ) and radius $\displaystyle \sqrt{\frac{\alpha^2 + \beta^2}{4}}$

x2 + y2 – αx – βy = 0

Note: Working rule to find the centre and radius of a circle whose equation is given.

Step I: Make the coefficients of x2 and y2 equal to 1 and right hand side equal to zero.

Step II: The coordinates of centre will be (α , β ) where α = – 1/2 (coefficient of x) and β = –1/2 (coefficient of y)

Step III: Radius $\displaystyle = \sqrt{\alpha^2 + \beta^2 – constant \; term}$

Problem : Find the center and the radius of the circles

(i) 3x2 + 3y2 − 8x − 10y + 3 = 0.

(ii) x2 + y2 + 2x sinθ + 2y cosθ − 8 = 0.

(iii) 2x2 + λxy + 2y2+ (λ − 4)x + 6y − 5 = 0, for some λ.

Solution: (i) We rewrite the given equation as

x2 + y2 − x − y + 1 = 0 => g = − 4/3, f = − 5/3, c = 1

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

$\displaystyle \frac{\sqrt{32}}{9} = \frac{4\sqrt{2}}{3} \; units$

(ii) x2 + y2 + 2x sin θ + 2y cos θ − 8 = 0.

Centre of this circle is (−sin θ, − cos θ)

Radius $\displaystyle = \sqrt{sin^2 \theta + cos^2 \theta + 8}$

$\displaystyle = \sqrt{1 + 8}$

(iii) 2x2 + λxy + 2y2 + (λ − 4)x + 6y − 5 = 0

rewrite the equation as ,

$\displaystyle x^2 + \frac{\lambda}{2} x y + y^2 + (\frac{\lambda – 4}{2}) x + 3 y -\frac{5}{2} = 0$ ….(i)

Since, there is no term of xy in the equation of circle

λ/2 = 0 ⇒ λ = 0

So, equation (i) reduces to

$\displaystyle x^2 + y^2 – 2 x + 3 y -\frac{5}{2} = 0$

Centre (1 , -3/2)

Radius $\displaystyle = \sqrt{1 + \frac{9}{4} + \frac{5}{2}}$

Radius $\displaystyle = \frac{\sqrt{23}}{2} \; units$

#### Intercepts made by a circle on the axis

(i) Length of the intercept made by the circle

x2 + y2 + 2gx + 2fy + c = 0 on

(a) x-axis = AB = 2√(g2 − c)

(b) y-axis = CD = 2√(f2 − c)

(ii) Intercepts are always positive.

(iii) If the circle touches x-axis, then |AB| = 0 ⇒ c = g2

(iv) If the circle touches y-axis, then |CD| = 0 ⇒ c = f2

(v) If the circle touches both the axes, then |CD| = 0 = |AB|

⇒ c = g2 = f2.