Introduction to Circles

DEFINITION

A circle is the locus of a point which moves in a plane such that its distance from a fixed point is always constant. The fixed point is called the centre of the circle and the constant distance, the radius of the circle.

STANDARD EQUATION OF A CIRCLE

The equation of a circle with the centre at (a, b) and radius r, is given by (x – a)² + (y – b)² = r² .If the centre of the circle is at the origin and the radius is r, then equation of circle is x² + y² = r²_.

EQUATION OF CIRCLE IN DIFFERENT CONDITIONS

CONDITION FOR THE GENERAL EQUATION OF SECOND DEGREE IN X AND Y TO REPRESENT A CIRCLE

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

$\bullet$ Coefficient of x²=coefficient of y² i.e. a=b

$\bullet$ Coefficient of xy=0 i.e. h=0.

The general equation of a circle is of the form x² + y² + 2gx + 2fy + c = 0 where g, f and c are constants. The given equation of circle in the standard form can be written as ${(x+g)}^2+{(y+f)}^2={\left(\sqrt{g^2+f^2-c}\right)}^2$. Hence the coordinates of its centre are (-g, -f) and radius is $\sqrt{g^2+f^2-c}$.

Case I: if $g^2+f^2-c$>0, a real circle is possible.

Case II: if $g^2+f^2-c$=0 , the circle is called a point circle.

Case III: if $g^2+f^2-c$< 0 no real circle is possible.

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 coordinate of the centre will be (a, b) where a = $-\frac{1}{2}$(coefficient of x) and b = $-\frac{1}{2}$(coefficient of y).

STEP III: Radius=$\sqrt{a^2+b^2-\text{constant term}}$

Illustration 1: Find the centre and radius of the circle

(A) 2x²+2y²-4x-5y-1=0.

(B) $3x^2+\lambda xy+3y^2+(\lambda -3)x+(\lambda +6)y-6=0$ for some $\lambda$.

Solution: (A) The given equation of circle can be written as x²+y²-2x-5/2y-1/2=0 $\Rightarrow$g=-1, f=-5/4 c=-1/2

Hence the centre is (1,5/4) and radius is $\sqrt{1+\frac{25}{16}+\frac{1}{2}}$.

(B) The given equation can be written as $x^2+y^2+\frac{\lambda }{3}xy+\frac{(\lambda -3)}{3}x+\frac{(\lambda +6)}{3}y-2=0$

Since there is no term of xy in the equation of circle $\Rightarrow \lambda =0.$

Hence the given equation reduces to x²+y²-x+2y-2=0. Centre is (1/2, -1) and radius is $\frac{\sqrt{13}}{2}\text{units}$

EQUATION OF A CIRCLE WHOSE END POINTS OF ANY DIAMETER IS GIVEN

Equation of the circle with points P(x₁, y₁) and Q(x₂, y₂) as extremities of a diameter is given by (x – x₁)(x – x₂) + (y – y₁)(y – y₂) = 0.

INTERCEPT MADE BY THE CIRCLE ON THE AXIS

Let the equation of circle be x² + y² + 2gx + 2fy + c = 0………(1)

X- INTERCEPT: Intercept made by the circle on the x-axis is called the X-intercept. The circle will intersect the x-axis where y = 0$\Rightarrow$x² + 2gx + c = 0 …(2)

Here three cases arises

Case I: If discriminant > 0 i.e. $g^2-c$>0 circle will intersect the axis at two distinct and real points let A (x₁, 0) and B (x₂, 0). Length of the intercept

$AB=\left|x_2-x_1\right|=\sqrt{{\left(x_1+x_2\right)}^2-4x_1{x}_2}=2\sqrt{g^2-c}$.

Case II: If discriminant = 0 i.e. g²=c. Then the circle will touch the x-axis. In this case length of the intercept made by the circle on the x-axis will be zero.

Case III: If discriminant < 0 i.e. $g^2-c$<0, in this case circle will neither touch nor intersect the x-axis

Y- INTERCEPT: Intercept made by the circle on the y-axis is called the Y-intercept. The circle will intersect the y-axis where x = 0$\Rightarrow$y² + 2fy + c = 0 …(2)

Again three cases arises

Case I: When discriminant > 0 i.e. $f^2-c$>0 circle will intersect the y-axis at two distinct and real points say A(0, y₁) and B(0, y₂). Length of the y-intercept

$AB=\left|y_2-y_1\right|=\sqrt{{\left(y_1+y_2\right)}^2-4y_1{y}_2}=2\sqrt{f^2-c}$

Case II: If discriminant = 0 i.e. $f^2-c$=0 Then the circle will touch the axis. In this case length of the intercept made by the circle on the x-axis will be zero.

Case III: If discriminant < 0, i.e. $f^2-c$<0 in this case circle will neither touch nor intersect the axes.

Illustration 2: Find equation of the circle touching y–axis at (0, 3) and making intercept of 8 units on the x–axis.

Solution: Let the equation of circle be x² + y² + 2gx + 2fy + c = 0

putting x = 0, we get y² + 2fy + c = 0 … (1)

As it touches y-axis at (0, 3), (1) must be of the form (y - 3)² = 0

$\Rightarrow$ y² - 6x + 9 = 0

Comparing, we get, f = - 3 and c = 9

Now, putting y = 0, we get x² + 2gx + c = 0

So, |x₁ - x₂| = $\sqrt{{(x_1+x_2)}^2-4x_1{x}_2}$ = 2$\sqrt{g^2-c}$ = 8

$\Rightarrow$ g² - c = 16 $\Rightarrow$ g² = 25 $\Rightarrow$ g = $\pm$ 5

so, the equation is x² + y² $\pm$ 10x - 6y + 9 = 0

PARAMETRIC EQUATION OF A CIRCLE

Let the equation of circle be (x – a)² + (y – b)² = r². Hence from the diagram it is clear that co-ordinates of any point on the circle can be taken as (a + r cos$\theta$, b + r sinq) where 0 $⩽$ $\theta$ < 2$\pi$ .

x = a + r cos$\theta$ and y = b + r sin$\theta$ is called the parameter equation of the circle.

POSITION OF A POINT WITH RESPECT TO A CIRCLE

Let the equation of the given circle be x²+y²+2gx+2fy+c=0 and the given point be P($\alpha$,$\beta$). Now, If the distance of the point P($\alpha$,$\beta$) from the centre of the circle(O) is greater than the radius of the circle i.e. $PO\sqrt{g^{\text{2}}+f^2-c}$, then the point will lie outside the circle.

$\Rightarrow$ $\alpha$² + $\beta$² + 2$\alpha$g + 2$\beta$f + c > 0, point P will lie outside the circle.

Similarly

If $\alpha$² + $\beta$² + 2$\alpha$g + 2$\beta$f + c = 0, point P will lie outside the circle.

If $\alpha$² + $\beta$² + 2$\alpha$g + 2$\beta$f + c <0 0, point P will lie outside the circle.