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 topics such as circle, equation of tangent, normal and chord of contact.
Equation of Circle 1. Standard Equation of Circle:
\(\begin{array}{l}{{x}^{2}}+{{y}^{2}}={{r}^{2}}\end{array} \)
centre (0, 0) and Radius (r)
2. Equation of circle in centre radius form:
\(\begin{array}{l}{{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}\end{array} \)
Centre (h, k), Radius = r
3. Equation of circle in General form:
\(\begin{array}{l}{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\end{array} \)
Where (–g, –f ) centre
r2 = g2 + f2 – c .
Radius =
\(\begin{array}{l}\sqrt{{{g}^{2}}+{{f}^{2}}-c}\end{array} \)
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
\(\begin{array}{l}\left| \begin{matrix} {{x}^{2}}+{{y}^{2}} & x & y & 1 \\ x_{1}^{2}+y_{1}^{2} & {{x}_{1}} & {{y}_{1}} & 1 \\ x_{2}^{2}+y_{2}^{2} & {{x}_{2}} & {{y}_{2}} & 1 \\ x_{3}^{2}+y_{3}^{2} & {{x}_{3}} & {{y}_{3}} & 1 \\ \end{matrix} \right|=0\end{array} \)
.
Area of circle = πr 2
Perimeter = 2πr, where r is the radius.
Equation of Circle under Different Conditions
Touches both axis with centre (a, a) and radius r = a
\(\begin{array}{l}{{(x-a)}^{2}}+{{(y-a)}^{2}}={{a}^{2}}\end{array} \)
Touches x-axis only, with centre (α, a)
\(\begin{array}{l}{{(x-\alpha )}^{2}}+{{(y-a)}^{2}}={{a}^{2}}\end{array} \)
Touches y –axis only at (a, β)
\(\begin{array}{l}{{(x-a)}^{2}}+{{(y- \beta)}^{2}}={{a}^{2}}\end{array} \)
\(\begin{array}{l}{{x}^{2}}+{{y}^{2}}-\alpha x-\beta x=0\end{array} \)
Parametric Equation of Circle Equation of circle = x2 + y2 = r2
X = r cos Ɵ
Y = r sin Ɵ
Squaring both side
x2 + y2 = (r2 cos Ɵ + r2 sin2Ɵ)
= 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.
R – radius
cp > R , {Point lie outside}
cp = R , {on the curve}
cp < R , {inside the curve}
Parametric Equation of a Circle – Video Lesson
Equation of Tangents and Normal The Equation of Tangents and Normal are explained below. Let the equation of circle be
x2 + y2 + 2gx + 2fy + c = 0
Tangent at point P(x1 , y1 )
Equation of Tangent of Circle
\(\begin{array}{l}xx_{1}^{{}}+y{{y}_{1}}+g(x+{{x}_{1}})+f(y+{{y}_{1}})+c=0\end{array} \)
Tangent having slope ‘m’
y = mx + C
where
\(\begin{array}{l}c=\pm \left( \sqrt{{{g}^{2}}+{{f}^{2}}-c} \right)\left( \sqrt{1+{{m}^{2}}} \right)\end{array} \)
,
\(\begin{array}{l}c=\pm \sqrt[r]{1+{{m}^{2}}}\end{array} \)
Pair of tangents from external point p (x1 , y1 )
T2 = ss1
Where
\(\begin{array}{l}T\equiv x{{x}_{1}}+y{{y}_{1}}+g(x+{{x}_{1}})+f(y+{{y}_{1}})\end{array} \)
,
\(\begin{array}{l}S={{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\end{array} \)
and
\(\begin{array}{l}{{S}_{1}}\equiv x{{{}_{1}}^{2}}+{{y}_{1}}^{2}+2g{{x}_{1}}+2f{{y}_{1}}+c=0\end{array} \)
Equation of normal at p(x1 , y1 ) to the circle
\(\begin{array}{l}S\equiv {{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\end{array} \)
is
\(\begin{array}{l}\frac{x-{{x}_{1}}}{{{x}_{1}}+g}=\frac{y-{{y}_{1}}}{{{y}_{1}}+f}\end{array} \)
Equation of chord
Equation of chord PQ
\(\begin{array}{l}wher{{e}_{\downarrow }}\,T={{S}_{1}}\end{array} \)
\(\begin{array}{l}T\equiv x{{x}_{1}}+y{{y}_{1}}+g(x+{{x}_{1}})+f(y+{{y}_{1}})+c=\end{array} \)
\(\begin{array}{l}{{S}_{1}}\equiv x_{1}^{2}+y_{1}^{2}+2g{{x}_{1}}+2f{{y}_{1}}+c=0\end{array} \)
Chord of contact
AB is called chord of contact. Equation of contact is T = 0
\(\begin{array}{l}x{{x}_{1}}+y{{y}_{1}}+g(x+{{x}_{1}})+f(y+{{y}_{1}})+c=0\end{array} \)
Radical axis to the two Circles Equation of Radical axis to the two circle S1 & S2
\(\begin{array}{l}{{S}_{1}}\equiv {{x}^{2}}+{{y}^{2}}+2{{g}_{1}}x+2{{f}_{1}}y+{{c}_{1}}=0\end{array} \)
and
\(\begin{array}{l}{{S}_{2}}={{x}^{2}}+{{y}^{2}}+2{{g}_{2}}x+2{{f}_{2}}y+{{c}_{2}}=0\end{array} \)
Equation of Radical axis is
S1 – S2 = 0
Family of circles S1 + λS2 = 0
Where ‘λ’ is parameter
Particular Cases of Circle
Problems On Circles Illustration 1: Find the centre and the radius of the circle
\(\begin{array}{l}3{{x}^{2}}+3{{y}^{2}}-8x-10y+3=0.\end{array} \)
Solution:
We rewrite the given equation as
\(\begin{array}{l}{{x}^{2}}+{{y}^{2}}-\frac{8}{3}x-\frac{10}{3}y+1=0\end{array} \)
\(\begin{array}{l}\Rightarrow g=-\frac{4}{3},f=-\frac{5}{3},c=1.\end{array} \)
Hence the centre is
\(\begin{array}{l}\left( \frac{4}{3},\frac{5}{3} \right)\end{array} \)
and the radius is
\(\begin{array}{l}\sqrt{\frac{16}{9}+\frac{25}{9}-1}=\sqrt{\frac{32}{9}}=\frac{4\sqrt{2}}{3}.\end{array} \)
Illustration 2: Find the equation of the circle with centre (1, 2) and which passes through the point (4, 6).
Solution:
The radius of the circle is
\(\begin{array}{l}\sqrt{{{\left( 4-1 \right)}^{2}}+{{\left( 6-2 \right)}^{2}}}=\sqrt{25}=5.\end{array} \)
Hence the equation of the circle is
\(\begin{array}{l}{{\left( x-1 \right)}^{2}}+{{\left( y-2 \right)}^{2}}=25\end{array} \)
\(\begin{array}{l}\Rightarrow {{x}^{2}}+{{y}^{2}}-2x-4y=20.\end{array} \)
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.
Solution:
The required equation of the circle is
\(\begin{array}{l}\left( x+4 \right)\left( x-12 \right)+\left( y-3 \right)\left( y+1 \right)=0.\end{array} \)
On the y-axis,
\(\begin{array}{l}x=0\Rightarrow -48+{{y}^{2}}-2y-3=0\end{array} \)
\(\begin{array}{l}\Rightarrow {{y}^{2}}-2y-51=0\Rightarrow y=1\pm \sqrt{52}\end{array} \)
Hence the length of intercept on the y-axis
\(\begin{array}{l}=2\sqrt{52}=4\sqrt{13}.\end{array} \)
Illustration 4: Find the equation of the circle passing through (1, 1), (2, -1) and (3, 2).
Solution:
Let the equation be
\(\begin{array}{l}{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0.\end{array} \)
Substituting the coordinates of the three given points, we get
\(\begin{array}{l}2g+2f+c=-2,\end{array} \)
\(\begin{array}{l}4g-2f+c=-5,\end{array} \)
\(\begin{array}{l}6g+4f+c=-13.\end{array} \)
Solving the above three equations, we obtain:
\(\begin{array}{l}f=-1/2;g=-5/2,c=4.\end{array} \)
Hence the equation of the circle is
\(\begin{array}{l}{{x}^{2}}+{{y}^{2}}-5x-y+4=0.\end{array} \)
Illustration 5: Find the equation of the circle whose centre is (3, 4) and which touches the line
\(\begin{array}{l}5x+12y=1.\end{array} \)
Solution:
Let r be the radius of the circle. Then
r = distance of the centre i.e. point (3, 4) from the line
\(\begin{array}{l}5x+12y=1\end{array} \)
\(\begin{array}{l}=\left| \frac{15+48-1}{\sqrt{25+144}} \right|=\frac{62}{13}.\end{array} \)
Hence the required equation of the circle is
\(\begin{array}{l}{{\left( x-3 \right)}^{2}}+{{\left( y-4 \right)}^{2}}={{\left( \frac{62}{13} \right)}^{2}}\end{array} \)
\(\begin{array}{l}\Rightarrow {{x}^{2}}+{{y}^{2}}-6x-8y+\frac{381}{169}=0.\end{array} \)
Illustration 6: Find the greatest distance of the point P(10, 7) from the circle
\(\begin{array}{l}{{x}^{2}}+{{y}^{2}}-4x-2y-20=0.\end{array} \)
Solution:
Since
\(\begin{array}{l}{{S}_{1}}={{10}^{2}}+{{7}^{2}}-4\times 10-2\times 7-20>0,\end{array} \)
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
\(\begin{array}{l}=\sqrt{{{\left( 10-2 \right)}^{2}}+{{\left( 7-1 \right)}^{2}}}=10\end{array} \)
and CB = radius
\(\begin{array}{l}=\sqrt{4+1+20}=5\end{array} \)
∴ PB = PC + CB
\(\begin{array}{l}=\left( 10+5 \right)=15\end{array} \)
Illustration 7: 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
\(\begin{array}{l}2x-y=2.\end{array} \)
Then find the equation of the circle.
Solution:
The line joining (4, 3) and (2, 1) is also along a diameter.
So, the centre is the point of intersection of the diameter
\(\begin{array}{l}2x-y=2\end{array} \)
and the line \(\begin{array}{l}y-3=\frac{3-1}{4-2}\left( x-4 \right)\end{array} \)
i.e., \(\begin{array}{l}x-y-1=0.\end{array} \)
Solving these, we get the centre as (1, 0).
Also, the radius = the distance between (1, 0) and (2, 1)
\(\begin{array}{l}=\sqrt{2}\end{array} \)
Hence, the equation of circle is
\(\begin{array}{l}{{\left( x-1 \right)}^{2}}+{{\left( y-0 \right)}^{2}}=2\end{array} \)
or \(\begin{array}{l}{{x}^{2}}+{{y}^{2}}-2x-1=0.\end{array} \)
Illustration 8: Find the length of the common chord of the circles
\(\begin{array}{l}{{x}^{2}}+{{y}^{2}}+2x+6y=0\end{array} \)
and \(\begin{array}{l}{{x}^{2}}+{{y}^{2}}-4x-2y-6=0\end{array} \)
Solution:
Equation of common chord is
\(\begin{array}{l}{{S}_{1}}-{{S}_{2}}=0\end{array} \)
or \(\begin{array}{l}6x+8y+6=0\end{array} \)
or \(\begin{array}{l}3x+4y+3=0.\end{array} \)
Centre of S1 is C1 (-1, -3) and its radius is
\(\begin{array}{l}\sqrt{1+9}=\sqrt{10}.\end{array} \)
.
Distance of C1 from the common chord
\(\begin{array}{l}=\frac{\left| -3-12+3 \right|}{\sqrt{9+16}}=\frac{12}{5}\end{array} \)
∴ Length of common chord
\(\begin{array}{l}=2\sqrt{{{\left( \sqrt{10} \right)}^{2}}-{{\left( \frac{12}{5} \right)}^{2}}}=2\sqrt{10-\frac{144}{25}}=2\sqrt{\frac{106}{25}}=\frac{2\sqrt{106}}{5}\end{array} \)
Illustration 9: 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}
Solution:
To make the curve
\(\begin{array}{l}{{x}^{2}}+{{y}^{2}}=4\end{array} \)
homogeneous with respect to line \(\begin{array}{l}x+y=a,\end{array} \)
we should have
\(\begin{array}{l}{{x}^{2}}+{{y}^{2}}-4{{\left( \frac{x+y}{a} \right)}^{2}}=0\end{array} \)
\(\begin{array}{l}\Rightarrow \,\,\,\,\,{{a}^{2}}\left( {{x}^{2}}+{{y}^{2}} \right)-4\left( {{x}^{2}}+{{y}^{2}}+2xy \right)=0\end{array} \)
\(\begin{array}{l}\Rightarrow \,\,\,\,\,{{x}^{2}}\left( {{a}^{2}}-4 \right)+{{y}^{2}}\left( {{a}^{2}}-4 \right)-8xy=0\end{array} \)
Since this is a pair of perpendicular straight lines, we have
∴
\(\begin{array}{l}{{a}^{2}}-4+{{a}^{2}}-4=0\end{array} \)
\(\begin{array}{l}\Rightarrow\end{array} \)
\(\begin{array}{l}{{a}^{2}}=4\end{array} \)
\(\begin{array}{l}\Rightarrow a=\pm 2\end{array} \)
Hence, the required set of a is {-2, 2}.
Hence, the correct answer is (a).
Circles – Important Topics
Circles – Class 11 Maths Chapter 11 Quiz
Circles – Important Questions
Circles – Solved Problems
Straight Lines and Circles – Top 12 Important Questions
Frequently Asked Questions Give the standard equation of a Circle with centre (0, 0) and radius, r.
The standard equation of a Circle with centre (0, 0) and radius, r is given by x2 + y2 = r2 .
What do you mean by a chord of a circle?
A chord is a line segment joining two points on the circle. The diameter is the largest chord.
Give the standard equation of a Circle with centre (h, k) and radius, r.
The standard equation of a Circle with centre (h, k) and radius, r is given by (x-h)2 + (y-k)2 = r2 .
