Circles

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:

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

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

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

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

• Touches y –axis only at (a, β)

• Passes through origin with centre $\left( \frac{\alpha }{2},\frac{\beta }{2} \right)$

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

Tangent having slope ‘m’

y = mx + C

where $c=\pm \left( \sqrt{{{g}^{2}}+{{f}^{2}}-c} \right)\left( \sqrt{1+{{m}^{2}}} \right)$,

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

T² = ss₁

Where $T\equiv x{{x}_{1}}+y{{y}_{1}}+g(x+{{x}_{1}})+f(y+{{y}_{1}})$,

Equation of normal at p(x₁, y₁) to the circle $S\equiv {{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ is

Equation of chord

Equation of chord PQ

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

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 $3{{x}^{2}}+3{{y}^{2}}-8x-10y+3=0.$

Solution:

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

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

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

Solution:

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

Hence the equation of the circle is

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

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

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

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

Solution:

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

Substituting the coordinates of the three given points, we get

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 $5x+12y=1.$

Solution:

Let r be the radius of the circle. Then

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

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

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

Solution:

Since ${{S}_{1}}={{10}^{2}}+{{7}^{2}}-4\times 10-2\times 7-20>0,$ 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 $=\sqrt{{{\left( 10-2 \right)}^{2}}+{{\left( 7-1 \right)}^{2}}}=10$

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

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

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 $2x-y=2.$ 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 $2x-y=2$ and the line $y-3=\frac{3-1}{4-2}\left( x-4 \right)$ i.e., $x-y-1=0.$

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

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

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

Illustration 8: Find the length of the common chord of the circles ${{x}^{2}}+{{y}^{2}}+2x+6y=0$ and ${{x}^{2}}+{{y}^{2}}-4x-2y-6=0$

Solution:

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

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

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

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

Illustration 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 ${{x}^{2}}+{{y}^{2}}=4$ homogeneous with respect to line $x+y=a,$ we should have

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

∴ ${{a}^{2}}-4+{{a}^{2}}-4=0$ $\Rightarrow$ ${{a}^{2}}=4$ $\Rightarrow a=\pm 2$

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

Hence, the correct answer is (a).