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Circle - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Circle(Definition) is considered one the most difficult concept.

  • 150 Questions around this concept.

Solve by difficulty

The equation of a circle with origin as a centre and passing through equilateral triangle whose median is of length 3a is

A square is inscribed in the circle \mathrm{x^{2}+y^{2}-2 x+4 y+3=0}. Its sides are parallel to the coordinate axes. Then one vertex of the square is:

The equation \mathrm{x^{2}+y^{2}+4 x+6 y+13=0} represents:

Concepts Covered - 1




A circle is the locus of a moving point such that its distance from a fixed point is constant.

The fixed point is called the centre (O) of the circle and constant distance is called its radius (r)

Equation of circle

Centre-Radius Form

The equation of a circle with centre at C (h,k) and radius r is (x - h)+ (y - k)= r2


Let P(x, y) be any point on the circle. Then, by definition, | CP | = r.

Using the distance formula, we have

\\ \mathrm{{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\sqrt{(x-h)^{2}+(y-k)^{2}}=r}} \\\\ \mathrm{i.e.\;\;\;\;\;\;\;\;\;\;\;\;\;\;{(x-h)^{2}+(y-k)^{2}=r^{2}}}

If the centre of the circle is the origin or (0,0) then equation of the circle becomes
\\(x-0)^{2}+(y-0)^{2}=r^{2}\\\text{i.e. }x^2+y^2=r^2


General Form:

The equation of a circle with centre at (h,k) and radius r is 

\\ {\Rightarrow(x-h)^{2}+(y-k)^{2}=r^{2}} \\ {\Rightarrow x^{2}+y^{2}-2 h x-2 k y+h^{2}+k^{2}-r^{2}=0\;\;\;\;\;\;\;\;\;\;\;\ldots(i)} \\ {\text { Which is of the form : }} \\ {\mathbf{x}^{2}+\mathbf{y}^{2}+2 \mathbf{g} \mathbf{x}+2 \mathbf{f} \mathbf{y}+\mathbf{c}=\mathbf{0}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots(ii)}

This is known as the general equation of the circle.

To get radius and centre if only the equation of the circle (ii) is given:

Compare eq (i) and eq (ii)

h = - g, k = - h  and c = h+ k- r2

Coordinates of the centre  (-g,-f)

Radius =\sqrt{ g^2+f^2-c }


Nature of the Circle

For the standard equation of a circle x2+y2+2gx+2fy+c=0 whose radius is given as \sqrt{ g^2+f^2-c }

Now the following cases arise

  1. If g2+f2-c > 0, then the radius of the circle will be real. Hence, the circle is a real circle.

  2. If g2+f2-c = 0,  then the radius of the circle will be real (=0). Hence, the circle is a Point circle because the radius is 0.

  3. If g2+f2-c < 0, then the radius of the circle will be imaginary. Hence, the circle is an imaginary circle.

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