Equation Of A Circle

Before deriving the equation of a circle, let us focus on Circle is a set of all points which are equally spaced from a fixed point in a plane.

  • The fixed point is called the centre of the circle.
  • The distance between centre and any point on the circumference is called the radius of the circle.

Equation of a Circle:

Centre is Origin:

Equation of a Circle

Consider an arbitrary point \(P(x,y)\) on the circle. Let \(a\) be the radius of the circle which is equal to \(OP\).

We know that, distance between the point \((x,y)\) and origin \((0,0)\)can be found using the distance formula which is equal to-

\(\sqrt{x^{2}+y^{2}} = a\)

Therefore, the equation of a circle, with centre as origin is,

\( x^{2}+y^{2} = a^{2}\)

Consider a circle whose centre is at the origin and radius is equal to 8 units.

For the given condition, the equation of a circle is given as

\(x^{2}+y^{2} = 8^{2}\)

Centre is not origin:

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

Therefore radius of a circle is CP.

By using distance formula,

\((x-h)^{2}+(y-k)^{2} = CP^{2}\)

Let radius be a.

Therefore, equation of the circle with centre \((h,k)\) and the radius \(a\) is,

\((x-h)^{2}+(y-k)^{2} = a^{2}\)

Example: Find the equation of the circle whose centre is \((3,5)\) and radius is 4 units.

Solution: Here, centre of the circle is not origin.

Therefore, the general equation of the circle is,

\((x-3)^2+(y-5)^2 \) = \(4^2\)

\(x^2-6x+9+y^2-10y+25\) = \(16\)

\(x^2+y^2-6x-10y+18\) = \(0\)

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

\(x^2~+~y^2~+~2gx~+~2fy~+~c\) = \(0\), for all values of \(g\),\(f\) and \(c\).

Adding \(g^2~+~f^2\) on both sides of the equation gives,

\(x^2~+~2gx~+~g^2~+~y^2~+~2fy~+~f^2\) = \(g^2~+~f^2~-~c\) ………………(1)

Since, \((x+g)^2 = x^2+2gx+g^2\) and \((y+f)^2 = y^2+2fy+f^2\) substituting the values in equation (1), we have

\((x+g)^2+(y+f)^2 = g^2+f^2-c\) …………….(2)

Comparing (2) with \((x-h)^2+(y-k)^2 = a^2\), where \((h,k)\) is the centre and \(a\) is the radius of the circle.

\(h =-g\), \(k = -f\)

\(a = g^{2}+ f^{2} -c\)


\(x^2+y^2+2gx+2fy+c\) = \(0\), represents the circle with centre \((-g,-f)\) and radius equal to \(a = g^{2}+ f^{2} -c\).

  • If \(g^2+f^2>c\), then the radius of the circle is real.
  • If \(g^2+f^2 = c\), then the radius of the circle is zero which tells us that the circle is a point which coincides with the centre. Such type of circle is called as point circle
  • \(g^2+f^2<c\), then the radius of the circle become imaginary. Therefore it is a circle having real centre and imaginary radius.

Example: Equation of a circle is \( x^2+y^2-12x-16y+19 = 0\). Find centre and radius of the circle.

Solution: Given equation is of the form \(x^2~+~y^2~+~2gx~+~2fy~+~c\) = \(0\),

\(2g\) = \(-12\), \(2f\) = \(-16\),\(c\) = \(19\)

\(g\) = \(-6\),\(f\) = \(-8\)

Centre of the circle is \((6,8)\)

Radius of the circle = \(√{(-6)^2~+~(-8)^2~-~19}\) = \(√{100~-~19}\) = \(9\) units.

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