Circles

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

$xx_{1}^{{}}+y{{y}_{1}}+g(x+{{x}_{1}})+f(y+{{y}_{1}})+c=0$

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

$S={{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ and

${{S}_{1}}\equiv x{{{}_{1}}^{2}}+{{y}_{1}}^{2}+2g{{x}_{1}}+2f{{y}_{1}}+c=0$

 

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

$\frac{x-{{x}_{1}}}{{{x}_{1}}+g}=\frac{y-{{y}_{1}}}{{{y}_{1}}+f}$

Equation of chord

 

Equation of chord PQ

$wher{{e}_{\downarrow }}\,T={{S}_{1}}$ $T\equiv x{{x}_{1}}+y{{y}_{1}}+g(x+{{x}_{1}})+f(y+{{y}_{1}})+c=$ ${{S}_{1}}\equiv x_{1}^{2}+y_{1}^{2}+2g{{x}_{1}}+2f{{y}_{1}}+c=0$

Chord of contact

 

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

$x{{x}_{1}}+y{{y}_{1}}+g(x+{{x}_{1}})+f(y+{{y}_{1}})+c=0$

Radical axis to the two Circles

Equation of Radical axis to the two circle S₁ & S₂

${{S}_{1}}\equiv {{x}^{2}}+{{y}^{2}}+2{{g}_{1}}x+2{{f}_{1}}y+{{c}_{1}}=0$ and

${{S}_{2}}={{x}^{2}}+{{y}^{2}}+2{{g}_{2}}x+2{{f}_{2}}y+{{c}_{2}}=0$

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

$\sqrt{\frac{16}{9}+\frac{25}{9}-1}=\sqrt{\frac{32}{9}}=\frac{4\sqrt{2}}{3}.$

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

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

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.

Solution:

The required equation of the circle is

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

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

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

Solving the above three equations, we obtain:

$f=-1/2;g=-5/2,c=4.$

Hence the equation of the circle is

${{x}^{2}}+{{y}^{2}}-5x-y+4=0.$

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

${{x}^{2}}+{{y}^{2}}-4{{\left( \frac{x+y}{a} \right)}^{2}}=0$ $\Rightarrow \,\,\,\,\,{{a}^{2}}\left( {{x}^{2}}+{{y}^{2}} \right)-4\left( {{x}^{2}}+{{y}^{2}}+2xy \right)=0$ $\Rightarrow \,\,\,\,\,{{x}^{2}}\left( {{a}^{2}}-4 \right)+{{y}^{2}}\left( {{a}^{2}}-4 \right)-8xy=0$

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

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

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

Hence, the correct answer is (a).