General Equation of the Circle | Standard Form of Circle

What is a Circle ?

A circle is the locus of a point which moves in a plane in such way that its distance from a fixed point(in the same given plane) remains constant. The fixed point is called center of circle and the constant distance is called radius of the circle. The general equation of the circle and standard form of circle are given below with examples.

General Equation of the Circle :

The general equation of the circle is

\(x^2 + y^2 + 2gx + 2fy + c\) = 0.

where g, f, c are constants and center is (-g, -f) and radius r = \(\sqrt{g^2 + f^2 – c}\).

(i)  If \({g^2 + f^2 – c}\) > 0, then r is real and positive and the circle is a real circle.

(ii) If \({g^2 + f^2 – c}\) = 0, then radius r = 0 and circle is a point circle.

(iii) If \({g^2 + f^2 – c}\) < 0, then r is imaginary and circle is also an imaginary circle with real center.

(iv) If \(x^2 + y^2 + 2gx + 2fy + c\) = 0, has three constants and to get the equation of the circle at least three conditions should be known \(\implies\) A unique circle passes through three non-collinear points.

Example : Find the center and radius of the circle \(3x^2 + 3y^2 – 8x -10y + 3\) = 0.

Solution : We rewrite the equation as, dividing whole equation by 3

\(x^2 + y^2 – {8\over 3}x – {10\over 3}y + 1\) = 0

Comparing it with general equation of the circle \(x^2 + y^2 + 2gx + 2fy + c\) = 0, we get

g = -\(4\over 3\), f = -\(5\over 3\), c = 1.

Hence center of the circle is (\(4\over 3\), \(5\over 3\))

and radius is \(\sqrt{{16\over 9}+{25\over 9}-1}\) = \(\sqrt{32\over 9}\).

General Second Degree Equation :

The general second degree in x and y, \(ax^2 + by^2 + 2hxy + 2gx + 2fy + c\) = 0 represents a circle if :

(i) coefficient of \(x^2\) = coefficient of \(y^2\) or a = b \(\ne\) 0.

(ii) coefficient of xy = 0 or h = 0.

(iii) \({g^2 + f^2 – c}\) \(\ge\) 0(for a real circle).

Standard Form of Circle :

If (h, k) is the center and r is the radius of the circle then standard form of circle equation is

\((x-h)^2 + (y-k)^2\) = \(r^2\).

(i)  If centre is origin (0,0) and radius is ‘r’ then the equation of circle is \(x^2\) + \(y^2\) = \(r^2\)

(ii)  If radius of circle is zero then the equation of circle is \((x-h)^2 + (y-k)^2\) = 0. Such a circle is called zero circle or point circle.

(iii)  When circle touches x-axis then equation of the circle is \((x-h)^2 + (y-k)^2\) = \(k^2\)

(iv)  When circle touches y-axis then the equation of circle is \((x-h)^2 + (y-k)^2\) = \(h^2\)

(v)  When circle touches both the axes(x-axis and y-axis) then equation of circle is \((x-h)^2 + (y-h)^2\) = \(h^2\)

(vi)  When circle passes through the origin and centre of the circle is (h,k) then radius \(\sqrt{h^2 + k^2}\) = r and intercept cut on x-axis is 2h and intercept cut on y-axis is 2k and equation of circle is \((x-h)^2 + (y-k)^2\) = \(h^2 + k^2\) or \(x^2\) + \(y^2\) – 2hx – 2ky = 0

Example : Find the center and radius of the circle \(x^2 + y^2\) = 9.

Solution : Comparing it with standard form of circle equation is \((x-h)^2 + (y-k)^2\) = \(r^2\). we get

h = 0, k = 0, r = 3.

Hence center of the circle is (0, 0).

and radius is 3.

Intercepts cut by the circle on axes

The intercepts cut by the circle \(x^2 + y^2 + 2gx + 2fy + c\) = 0 on :

(i)  x-axis = 2\(\sqrt{g^2 – c}\)

(ii)  y-axis = 2\(\sqrt{f^2 – c}\)

Hope you learnt general equation of the circle and standard form of circle, learn more concepts of circle and practice more questions to get ahead in the competition. Good luck!

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