CIRCLE-10 (Limiting Point)

Limiting Point

Limiting points of a system of co-axial circles are the centres of the point circles belonging to the family (circles whose radii are zero are called point circle)

1. Limiting points of the co-axial system

Let the circle is x2+y2+2gx+c=0

Where g is a variable and c is constant

centre (g,0) and radius g2c respectively

Let g2c=0 radius

g2c=0

g2=c

g=±c

Thus we get the two limiting points of the given co-axial system as Misplaced &

The limiting points are real and distinct, real and coincident or imaginary according as C>,=,<0.

2. System of co-axial circles whose limiting points are given

Let (α,β) and (γ,δ) be the two given limiting points

Then corresponding circles with zero radii are

(xα)2+(yβ)2=0=x2+y22αx2βy+α2+β2=0

(xγ)2+(yδ)2=0=x2+y22γx2δy+γ2+δ2=0

System of co-axial circle equation is

x2+y22αx2βy+α2+β2+λ(x2+y22γx+2δy+γ2+δ2)=0(λ1)

centre of this circle is (α+γλ1+λ,β+δλ1+λ)

and radius =(α+γλ1+λ)2+(β+δλ1+λ)2(α2+β2+λγ2+λδ2)1+λ=0

After solving find λ substitute in (1)

We get the limiting point of co-axial system.

Properties of Limiting points

1. The limiting point of a system of co-axial circles are conjugate points with respect to any member of the system.

Let the equation of any circle be

x2+y2+2gx+c=0

limiting points of (1) are (c,0),(c,0)

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The polar of the point (c,0) is

xc+g(x+c)+c=0

(x+c)(g+c)=0

x+c=0x=c

(c,0) ) lies on this.

Similarly (c,0) lies on polar with respect to (c,0)

These limiting points (c,0) and (c,0) are conjugate points

2. Every circle through the limiting points of a co-axial system is orthogonal to all circles of the system

Let the equation of any circle x2+y2+2gx+c=0 where g is a variable and c is a constant Limiting point of this circle are (c,0) and (c,0)

Now let, x2+y2+2 g1x+2f1y+c1= be any circle passing through the limiting points (c,0) and (c,0).

c2 g1c+c1=0 (3) and c+2 g1c+c1=0

Solving (3) and (4). We get g1=0 and c1=c

The equation becomes

x2+y2+2f1yc=0

applying condition of orthogonality

2gg1+2ff1=c+c1 o+o=cc

Hence condition is satisfied for all values of g1 and f1

Examples

1. If the origin be one limiting point of a system of co-axial circles of which x2+y2+3x+4y+25=0 is a member, find the other limiting point.

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Solution :

Equation of circle with origin (0,0) as limiting point is x2+y2=0

Given that one member of system of co-axial circle is x2+y2+3x+4y+25=0

The system of co-axial circles is

x2+y2+31+λx+41+λy+251+λ=0

centre (32(1+λ),21+λ)

radius 94(1+λ)2+4(1+λ)225(1+λ)=0

254(1+λ)2251+λ=0

14(1+λ)=0

1+λ=1/4

λ=1/41=3/4

centre (6,8) is the other limiting point of the system.

2. Find the radical axis of co-axial system of circles whose limiting points are (1,2) and (2,3).

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Solution :

Equations of circles with limiting points (1,2) and (2,3) are

(x+1)2+(y2)2=0,x2+y2+2x4y+5=0

(x2)2+(y3)2=0,x2+y24x6y+13=0

respectively

Equation of radical axis of (1) and (2) is

S1S2=0

6x+2y8=0

3x+y4=0

3. Find the equation of the circle which passes through the origin and belongs to the co-axial of circles whose limiting points are (1,2) and (4,3)

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Solution :

Equation of circles with limiting points (1,2) and (4,3) are

(x1)2+(y2)2=0x2+y22x4y+5=0

(x4)2+(y3)2=0x2+y28x6y+25=0

System of co-axial of circles equation is

x2+y22x4y+5+λ(x2+y28x6y+25)=0

equation (1) passes through origin

5+25λ=0

λ=1/5

Substituting in (1) we get

4(x2+y2)2x14y=0

2x2+2y2x7y=0