2. Standard equations, Properties and Application of a circle

Standard equations , Properties and Application of a circle

Circle:

Definition: A circle is the set of all points in a plane that are equidistant from a fixed

point in the plane.

The fixed point is called the centre of the circle and the distance from the centre

to a point on the circle is called the radius of the circle.

Given C (h, k) be the centre and r the radius of circle. Let P(x, y) be any point on

the circle . Then, by the definition, | CP | = r . By the distance formula,

we have

(x-h)² + (y-k)²= r²

This is the required equation of the circle with centre at (h,k) and radius r .

General form of Equation of a Circle

The general equation of any type of circle is represented by:

x² + y² + 2gx + 2fy + c = 0, for all values of g, f and c.

Adding g² + f² on both sides of the equation gives,

x² + 2gx + g²+ y² + 2fy + f²= g² + f² − c ………………(1)

Since, (x+g)² = x²+ 2gx + g² and (y+f)² =y² + 2fy + f² substituting the values in equation (1), we have

(x+g)²+ (y+f)² = g² + f²−c …………….(2)

Comparing (2) with (x−h)² + (y−k)² = a², where (h, k) is the center and ‘a’ is the radius of the circle.

h=−g, k=−f

a² = g²+ f²−c

Therefore,

x² + y² + 2gx + 2fy + c = 0, represents the circle with centre (−g,−f) and radius equal to a² = g² + f²− c.

• If g² + f² > c, then the radius of the circle is real.
• If g² + f² = c, then the radius of the circle is zero which tells us that the circle is a point that coincides with the center. Such a type of circle is called a point circle
• g² + f² <c, then the radius of the circle become imaginary. Therefore, it is a circle having a real center and imaginary radius.

 

N.B.:  Standard Equation of Circle:

centre (0, 0) and Radius (r)

 Equation of circle in centre radius form:

Centre (h, k), Radius = r

 Equation of circle in General form:

Where (–g, –f ) centre

r² = g² + f² – c .

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

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

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

Area of circle = πr²

Perimeter = 2πr, where r is the radius.

Example: Find an equation of the circle with centre at (0,0) and radius r.

Solution :Here h = k = 0. Therefore, the equation of the circle is x² + y² = r²

Example: Find the equation of the circle with centre (–3, 2) and radius 4.

Solution: Here h = –3, k = 2 and r = 4. Therefore, the equation of the required circle is

(x + 3) ² + (y –2)² = 16

Example  : Find the centre and the radius of the circle x² + y² + 8x + 10y – 8 = 0

Solution : The given equation is

(x² + 8x) + (y2²+ 10y) = 8

Now, completing the squares within the parenthesis, we get

(x²+ 8x + 16) + (y² + 10y + 25) = 8 + 16 + 25

i.e. (x + 4)² + (y + 5)² = 49

i.e. {x – (– 4)} ²+ {y – (–5)} ² = 7²

Therefore, the given circle has centre at (– 4, –5) and radius 7.

Example : Find the equation of the circle which passes through the points (2, – 2), and (3, 4) and whose centre lies on the line x + y = 2.

Solution: Let the equation of the circle be (x – h)² + (y – k)² = r².

Given that the circle passes through the points (2, –2) and (3, 4).

Thus,

(2 – h)² + (–2 – k)² = r²….(1)

and (3 – h)² + (4 – k)² = r²….(2)

Also, given that the centre lies on the line x + y = 2.

⇒ h + k = 2 ….(3)

Solving the equations (1), (2) and (3), we get

h = 0.7, k = 1.3 and r² = 12.58

Hence, the equation of the required circle is

(x – 0.7)² + (y – 1.3)² = 12.58