Circles

A circle is the locus of paints which moves in a plane such that its distance from a fixed paint 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 = \(\sqrt{{{g}^{2}}+{{f}^{2}}-c}\)

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

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, b)

• 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

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 \(c=\pm \left( \sqrt{{{g}^{2}}+{{f}^{2}}-c} \right)\left( \sqrt{1+{{m}^{2}}} \right)\) \(c=\pm \sqrt[r]{1+{{m}^{2}}}\)

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

T² = ss₁

Where \(T\equiv x{{x}_{1}}+y{{y}_{1}}+g(x+{{x}_{1}})+f(y+{{y}_{1}})\) \(S={{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\) \({{S}_{1}}\equiv x{{{}_{1}}^{2}}+{{y}_{1}}^{2}+2g{{x}_{1}}+2f{{y}_{1}}+c=0\)

Equation of normal at p(x₁, y₁)

Let circle be

P(x₁, y₁)

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

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

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

Solution:

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

or \(x-2y-5=0\)

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

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

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

Solution:

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

If \(\frac{\left| -1-7m \right|}{\sqrt{1+{{m}^{2}}}}=5\) \(\Rightarrow m=\frac{3}{4},-\frac{4}{3}\)

The product of both the slopes is -1.

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

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

Solution:

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 \(x+2y=3.\)

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 \(2x-y=2.\) 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 11: 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 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}

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

Hence, the correct answer is (a).