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’ while the fixed distance is called the ‘radius’ . This article helps you to have a clear idea of the circle, equation of tangent, normal and chord of contact. 

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.

Area of circle = πr2

Perimeter = 2πr, where r is the radius.

Equation of Circle under Different Conditions

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


Equation of circle touching both axis



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

Equation of circle touching x axis


(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)


Equation of circle passes through origin with center

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)


Parametric equation of a circle
= r2 (cos2 Ɵ + sin2 Ɵ)

x2 + y2 = r2

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

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

Position of a point with respect to circle


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 tangent of circle


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


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

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

Family of circles


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).