Conic sections have their unique physical definitions and unique algebraic formulas, relating characteristics of points that are on them. In the circle, there is a simple centre and every point on the circumference of the circle is a certain distance away from the centre. In short, a circle is the locus of points 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 while the fixed distance is called the ‘radius’ of the circle.

Equation of Circle

1. Standard Equation of Circle:


centre (0, 0) and Radius (r)

2. Equation of circle in centre radius form:


Centre (h, k), Radius = r

3. Equation of circle in General form:


Where (–g, –f ) centre

r2 = g2 + f2 – c .

Radius = g2+f2c\sqrt{{{g}^{2}}+{{f}^{2}}-c}

4. 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

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

x2+y2xy1x12+y12x1y11x22+y22x2y21x32+y32x3y31=0\left| \begin{matrix} {{x}^{2}}+{{y}^{2}} & x & y & 1 \\ x_{1}^{2}+y_{1}^{2} & {{x}_{1}} & {{y}_{1}} & 1 \\ x_{2}^{2}+y_{2}^{2} & {{x}_{2}} & {{y}_{2}} & 1 \\ x_{3}^{2}+y_{3}^{2} & {{x}_{3}} & {{y}_{3}} & 1 \\ \end{matrix} \right|=0.

Equation of Circle under Different Conditions

  • Touches both axis with centre (a, a) and radius r = a


Touches x-axis only, with centre (α, a)

(xα)2+(ya)2=a2{{(x-\alpha )}^{2}}+{{(y-a)}^{2}}={{a}^{2}}
  • Touches y –axis only at (a, β)

(xa)2+(yβ)2=a2{{(x-a)}^{2}}+{{(y- \beta)}^{2}}={{a}^{2}}
  • Passes through origin with centre (α2,β2)\left( \frac{\alpha }{2},\frac{\beta }{2} \right)

x2+y2αxβx=0{{x}^{2}}+{{y}^{2}}-\alpha x-\beta x=0

Parametric Equation of Circle

Equation of circle = x2 + y2 = r2

X = r cos

Y = r sin

Squaring both side

x2 + y2 = (r2 cos Ɵ + r2 sin2Ɵ)

= r2 (cos2 Ɵ + sin2 Ɵ)

x2 + y2 = r2

Position of a paint w.r.t. to circle

Let the circle be x2 + y2 + 2gx + 2fy + c = 0 and p(x1, y1) be the point.

R – radius

cp > R , {Point lie outside}

cp = R , {on the curve}

cp < R , {inside the curve}

Equation of Tangents and Normal

The Equation of Tangents and Normal are explained below. Let the equation of circle be

x2 + y2 + 2gx + 2ky + c = 0

Tangent at point P(x1, y1)

Equation of Tangent of Circle


Tangent having slope ‘m’

y = mx + C

where c=±(g2+f2c)(1+m2)c=\pm \left( \sqrt{{{g}^{2}}+{{f}^{2}}-c} \right)\left( \sqrt{1+{{m}^{2}}} \right) c=±1+m2rc=\pm \sqrt[r]{1+{{m}^{2}}}

Pair of tangents from external point p (x1, y1)

T2 = ss1

Where Txx1+yy1+g(x+x1)+f(y+y1)T\equiv x{{x}_{1}}+y{{y}_{1}}+g(x+{{x}_{1}})+f(y+{{y}_{1}}) S=x2+y2+2gx+2fy+c=0S={{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0 S1x12+y12+2gx1+2fy1+c=0{{S}_{1}}\equiv x{{{}_{1}}^{2}}+{{y}_{1}}^{2}+2g{{x}_{1}}+2f{{y}_{1}}+c=0

Equation of normal at p(x1, y1)

Let circle be

Sx2+y2+2gx+2fy+c=0S\equiv {{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0

P(x1, y1)

So, equation of normal xx1x1+g=yy1y1+f\frac{x-{{x}_{1}}}{{{x}_{1}}+g}=\frac{y-{{y}_{1}}}{{{y}_{1}}+f}

Equation of chord

Equation of chord PQ

whereT=S1wher{{e}_{\downarrow }}\,T={{S}_{1}} Txx1+yy1+g(x+x1)+f(y+y1)+c=T\equiv x{{x}_{1}}+y{{y}_{1}}+g(x+{{x}_{1}})+f(y+{{y}_{1}})+c= S1x12+y12+2gx1+2fy1+c=0{{S}_{1}}\equiv x_{1}^{2}+y_{1}^{2}+2g{{x}_{1}}+2f{{y}_{1}}+c=0

Chord of contact

AB is called chord of contact. Equation of contact is T = 0


Radical axis to the two Circles

Equation of Radical axis to the two circle S1 & S2

S1x2+y2+2g1x+2f1y+c1=0{{S}_{1}}\equiv {{x}^{2}}+{{y}^{2}}+2{{g}_{1}}x+2{{f}_{1}}y+{{c}_{1}}=0 and


Equation of Radical axis is

S1 – S2 = 0

Family of circles

S1 + λS2 = 0

Where ‘λ’ is parameter

Problems On Circles

Illustration 1: Find the centre and the radius of the circle 3x2+3y28x10y+3=0.3{{x}^{2}}+3{{y}^{2}}-8x-10y+3=0.


We rewrite the given equation as x2+y283x103y+1=0{{x}^{2}}+{{y}^{2}}-\frac{8}{3}x-\frac{10}{3}y+1=0 g=43,f=53,c=1.\Rightarrow g=-\frac{4}{3},f=-\frac{5}{3},c=1.

Hence the centre is (43,53)\left( \frac{4}{3},\frac{5}{3} \right) and the radius is



Illustration 2: Find the equation of the circle with centre (1, 2) and which passes through the point (4, 6).


The radius of the circle is (41)2+(62)2=25=5.\sqrt{{{\left( 4-1 \right)}^{2}}+{{\left( 6-2 \right)}^{2}}}=\sqrt{25}=5.

Hence the equation of the circle is

(x1)2+(y2)2=25{{\left( x-1 \right)}^{2}}+{{\left( y-2 \right)}^{2}}=25 x2+y22x4y=20.\Rightarrow {{x}^{2}}+{{y}^{2}}-2x-4y=20.


Illustration 3: Find the equation of the circle whose diameter is the line joining the points (-4, 3) and (12, -1). Find also the length of intercept made by it on the y-axis.


The required equation of the circle is

(x+4)(x12)+(y3)(y+1)=0.\left( x+4 \right)\left( x-12 \right)+\left( y-3 \right)\left( y+1 \right)=0.

On the y-axis, x=048+y22y3=0x=0\Rightarrow -48+{{y}^{2}}-2y-3=0 y22y51=0y=1±52\Rightarrow {{y}^{2}}-2y-51=0\Rightarrow y=1\pm \sqrt{52}

Hence the length of intercept on the y-axis =252=413.=2\sqrt{52}=4\sqrt{13}.


Illustration 4: Find the equation of the circle passing through (1, 1), (2, -1) and (3, 2).


Let the equation be x2+y2+2gx+2fy+c=0.{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0.

Substituting the coordinates of the three given points, we get

2g+2f+c=2,2g+2f+c=-2, 4g2f+c=5,4g-2f+c=-5, 6g+4f+c=13.6g+4f+c=-13.

Solving the above three equations, we obtain:


Hence the equation of the circle is



Illustration 5: Find the equation of the circle whose centre is (3, 4) and which touches the line 5x+12y=1.5x+12y=1.


Let r be the radius of the circle. Then

r = distance of the centre i.e. point (3, 4) from the line 5x+12y=15x+12y=1 =15+48125+144=6213.=\left| \frac{15+48-1}{\sqrt{25+144}} \right|=\frac{62}{13}.

Hence the required equation of the circle is (x3)2+(y4)2=(6213)2{{\left( x-3 \right)}^{2}}+{{\left( y-4 \right)}^{2}}={{\left( \frac{62}{13} \right)}^{2}} x2+y26x8y+381169=0.\Rightarrow {{x}^{2}}+{{y}^{2}}-6x-8y+\frac{381}{169}=0.


Illustration 6: Find the greatest distance of the point P(10, 7) from the circle x2+y24x2y20=0.{{x}^{2}}+{{y}^{2}}-4x-2y-20=0.


Since S1=102+724×102×720>0,{{S}_{1}}={{10}^{2}}+{{7}^{2}}-4\times 10-2\times 7-20>0, P lies outside the circle.

Join P with the centre C(2, 1) of the given circle.

Suppose PC cuts the circle at A and B, where A is nearer to C.

Then, PB is the greatest distance of P from the circle.

We have, PC =(102)2+(71)2=10=\sqrt{{{\left( 10-2 \right)}^{2}}+{{\left( 7-1 \right)}^{2}}}=10

and CB = radius =4+1+20=5=\sqrt{4+1+20}=5

∴ PB = PC + CB =(10+5)=15=\left( 10+5 \right)=15


Illustration 7: The tangent to circle x2+y2=5{{x}^{2}}+{{y}^{2}}=5 at (1, -2) also touches the circle x2+y28x+6y+20=0.{{x}^{2}}+{{y}^{2}}-8x+6y+20=0. Find the coordinates of the corresponding point of contact.


Equation of tangent to x2+y2=5{{x}^{2}}+{{y}^{2}}=5 at (1, -2) is

x(1)+y(2)5=0x\left( 1 \right)+y\left( -2 \right)-5=0

or x2y5=0x-2y-5=0

putting x=2y+5x=2y+5 in second circle, we get

(2y+5)2+y28(2y+5)+6y+20=0{{\left( 2y+5 \right)}^{2}}+{{y}^{2}}-8\left( 2y+5 \right)+6y+20=0 y2+2y+1=0\Rightarrow \,\,{{y}^{2}}+2y+1=0 y=1\Rightarrow \,\,y=-1 x=2+5=3\Rightarrow \,\,x=-2+5=3

Thus, point of contact is (3, -1).


Illustration 8: Find the angle between the two tangents from the origin to the circle (x7)2+(y+1)2=25.{{\left( x-7 \right)}^{2}}+{{\left( y+1 \right)}^{2}}=25.


Any line through (0, 0) be ymx=0y-mx=0 and it is a tangent to circle (x7)2+(y+1)2=25.{{\left( x-7 \right)}^{2}}+{{\left( y+1 \right)}^{2}}=25.

If 17m1+m2=5\frac{\left| -1-7m \right|}{\sqrt{1+{{m}^{2}}}}=5 m=34,43\Rightarrow m=\frac{3}{4},-\frac{4}{3}

The product of both the slopes is -1.

Hence, the angle between the two tangents is π/2.\pi /2.


Illustration 9: Find the equation of normal to the circle 2x2+2y22x5y+3=02{{x}^{2}}+2{{y}^{2}}-2x-5y+3=0 at (1, 1).


The centre of the circle is (1/2, 5/4)

Normal to circle at point (1, 1) is line passing through the points (1, 1) and (1/2, 5/4) which is x+2y=3.x+2y=3.


Illustration 10: A foot of the normal from the point (4, 3) to a circle is (2, 1) and a diameter of the circle has the equation 2xy=2.2x-y=2. Then find the equation of the circle.


The line joining (4, 3) and (2, 1) is also along a diameter.

So, the centre is the point of intersection of the diameter 2xy=22x-y=2 and the line y3=3142(x4)y-3=\frac{3-1}{4-2}\left( x-4 \right) i.e., xy1=0.x-y-1=0.

Solving these, we get the centre as (1, 0).

Also, the radius = the distance between (1, 0) and (2, 1) =2=\sqrt{2}

Hence, the equation of circle is (x1)2+(y0)2=2{{\left( x-1 \right)}^{2}}+{{\left( y-0 \right)}^{2}}=2or x2+y22x1=0.{{x}^{2}}+{{y}^{2}}-2x-1=0.


Illustration 11: Find the length of the common chord of the circles x2+y2+2x+6y=0{{x}^{2}}+{{y}^{2}}+2x+6y=0 and x2+y24x2y6=0{{x}^{2}}+{{y}^{2}}-4x-2y-6=0


Equation of common chord is S1S2=0{{S}_{1}}-{{S}_{2}}=0 or 6x+8y+6=06x+8y+6=0 or 3x+4y+3=0.3x+4y+3=0.

Centre of S1 is C1(-1, -3) and its radius is 1+9=10.\sqrt{1+9}=\sqrt{10}..

Distance of C1 from the common chord =312+39+16=125=\frac{\left| -3-12+3 \right|}{\sqrt{9+16}}=\frac{12}{5}

∴ Length of common chord =2(10)2(125)2=21014425=210625=21065=2\sqrt{{{\left( \sqrt{10} \right)}^{2}}-{{\left( \frac{12}{5} \right)}^{2}}}=2\sqrt{10-\frac{144}{25}}=2\sqrt{\frac{106}{25}}=\frac{2\sqrt{106}}{5}


Illustration 12: A pair of perpendicular straight lines pass through the origin and also through the point of intersection of the curve with The set containing the value of a is

(a) {-2, 2} (b) {-3, 3} (c) {-4, 4} (d) {-5, 5}


To make the curve x2+y2=4{{x}^{2}}+{{y}^{2}}=4 homogeneous with respect to line x+y=a,x+y=a, we should have

x2+y24(x+ya)2=0{{x}^{2}}+{{y}^{2}}-4{{\left( \frac{x+y}{a} \right)}^{2}}=0 a2(x2+y2)4(x2+y2+2xy)=0\Rightarrow \,\,\,\,\,{{a}^{2}}\left( {{x}^{2}}+{{y}^{2}} \right)-4\left( {{x}^{2}}+{{y}^{2}}+2xy \right)=0 x2(a24)+y2(a24)8xy=0\Rightarrow \,\,\,\,\,{{x}^{2}}\left( {{a}^{2}}-4 \right)+{{y}^{2}}\left( {{a}^{2}}-4 \right)-8xy=0

Since this is a pair of perpendicular straight lines, we have

a24+a24=0{{a}^{2}}-4+{{a}^{2}}-4=0 \Rightarrow a2=4{{a}^{2}}=4 a=±2\Rightarrow a=\pm 2

Hence, required set of a is {-2, 2}.

Hence, the correct answer is (a).