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 $\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{gx}+\mathrm{c}=0$

Where $\mathrm{g}$ is a variable and $\mathrm{c}$ is constant

$\therefore$ centre $(-\mathrm{g}, 0)$ and radius $\sqrt{\mathrm{g}^{2}-\mathrm{c}}$ respectively

Let $\sqrt{\mathrm{g}^{2}-\mathrm{c}}=0$ radius

$\mathrm{g}^{2}-\mathrm{c}=0$

$\mathrm{g}^{2}=\mathrm{c}$

$\mathrm{g}= \pm \sqrt{\mathrm{c}}$

Thus we get the two limiting points of the given co-axial system as $(\sqrt{\mathrm{c}}, 0) &(-\sqrt{\mathrm{c}}, 0)$

The limiting points are real and distinct, real and coincident or imaginary according as $\mathrm{C}>,=,<0$.

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

Let $(\alpha, \beta)$ and $(\gamma, \delta)$ be the two given limiting points

Then corresponding circles with zero radii are

$(x-\alpha)^{2}+(y-\beta)^{2}=0=x^{2}+y^{2}-2 \alpha x-2 \beta y+\alpha^{2}+\beta^{2}=0$

$(\mathrm{x}-\gamma)^{2}+(\mathrm{y}-\delta)^{2}=0=\mathrm{x}^{2}+\mathrm{y}^{2}-2 \gamma \mathrm{x}-2 \delta \mathrm{y}+\gamma^{2}+\delta^{2}=0$

System of co-axial circle equation is

$x^{2}+y^{2}-2 \alpha x-2 \beta y+\alpha^{2}+\beta^{2}+\lambda\left(x^{2}+y^{2}-2 \gamma x+2 \delta y+\gamma^{2}+\delta^{2}\right)=0 \quad(\lambda \neq-1)$

centre of this circle is $\left(\frac{\alpha+\gamma \lambda}{1+\lambda}, \frac{\beta+\delta \lambda}{1+\lambda}\right)$

and radius $=\sqrt{\left(\frac{\alpha+\gamma \lambda}{1+\lambda}\right)^{2}+\left(\frac{\beta+\delta \lambda}{1+\lambda}\right)^{2}-\frac{\left(\alpha^{2}+\beta^{2}+\lambda \gamma^{2}+\lambda \delta^{2}\right)}{1+\lambda}}=0$

After solving find $\lambda$ 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

$\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{gx}+\mathrm{c}=0$

limiting points of (1) are $(\sqrt{\mathrm{c}}, 0),(-\sqrt{\mathrm{c}}, 0)$

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The polar of the point $(\sqrt{\mathrm{c}}, 0)$ is

$\mathrm{x} \sqrt{\mathrm{c}}+\mathrm{g}(\mathrm{x}+\sqrt{\mathrm{c}})+\mathrm{c}=0$

$(\mathrm{x}+\sqrt{\mathrm{c}})(\mathrm{g}+\sqrt{\mathrm{c}})=0$

$\mathrm{x}+\sqrt{\mathrm{c}}=0 \Rightarrow \mathrm{x}=-\sqrt{\mathrm{c}}$

$\therefore(-\sqrt{c}, 0)$ ) lies on this.

Similarly $(\sqrt{\mathrm{c}}, 0)$ lies on polar with respect to $(-\sqrt{\mathrm{c}}, 0)$

$\therefore$ These limiting points $(-\sqrt{\mathrm{c}}, 0)$ and $(\sqrt{\mathrm{c}}, 0)$ 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 $\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{gx}+\mathrm{c}=0$ where $\mathrm{g}$ is a variable and $\mathrm{c}$ is a constant Limiting point of this circle are $(-\sqrt{\mathrm{c}}, 0)$ and $(\sqrt{\mathrm{c}}, 0)$

Now let, $\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{~g} _{1} \mathrm{x}+2 \mathrm{f} _{1} \mathrm{y}+\mathrm{c} _{1}=$ be any circle passing through the limiting points $(-\sqrt{\mathrm{c}}, 0)$ and $(\sqrt{\mathrm{c}}, 0)$.

$\therefore \mathrm{c}-2 \mathrm{~g} _{1} \sqrt{\mathrm{c}}+\mathrm{c} _{1}=0$ (3) and $\mathrm{c}+2 \mathrm{~g} _{1} \sqrt{\mathrm{c}}+\mathrm{c} _{1}=0$

Solving (3) and (4). We get $\mathrm{g} _{1}=0$ and $\mathrm{c} _{1}=-\mathrm{c}$

$\therefore$ The equation becomes

$ \mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{f} _{1} \mathrm{y}-\mathrm{c}=0 $

applying condition of orthogonality

$2 \mathrm{gg} _{1}+2 \mathrm{ff} _{1}=\mathrm{c}+\mathrm{c} _{1}$ $\mathrm{o}+\mathrm{o}=\mathrm{c}-\mathrm{c}$

Hence condition is satisfied for all values of $g _{1}$ and $f _{1}$

Examples

1. If the origin be one limiting point of a system of co-axial circles of which $x^{2}+y^{2}+3 x+4 y+25=0$ is a member, find the other limiting point.

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Solution :

Equation of circle with origin $(0,0)$ as limiting point is $x^{2}+y^{2}=0$

Given that one member of system of co-axial circle is $x^{2}+y^{2}+3 x+4 y+25=0$

$\therefore$ The system of co-axial circles is

$\mathrm{x}^{2}+\mathrm{y}^{2}+\frac{3}{1+\lambda} \mathrm{x}+\frac{4}{1+\lambda} \mathrm{y}+\frac{25}{1+\lambda}=0$

centre $\left(\frac{-3}{2(1+\lambda)}, \frac{-2}{1+\lambda}\right)$

radius $\frac{9}{4(1+\lambda)^{2}}+\frac{4}{(1+\lambda)^{2}}-\frac{25}{(1+\lambda)}=0$

$\frac{25}{4(1+\lambda)^{2}}-\frac{25}{1+\lambda}=0$

$1-4(1+\lambda)=0$

$1+\lambda=1 / 4$

$\lambda=1 / 4-1=-3 / 4$

$\therefore$ 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

$(\mathrm{x}+1)^{2}+(\mathrm{y}-2)^{2}=0, \mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{x}-4 \mathrm{y}+5=0$

$(x-2)^{2}+(y-3)^{2}=0, x^{2}+y^{2}-4 x-6 y+13=0$

respectively

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

$\mathrm{S} _{1}-\mathrm{S} _{2}=0$

$6 x+2 y-8=0$

$3 x+y-4=0$

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

$(\mathrm{x}-1)^{2}+(\mathrm{y}-2)^{2}=0 \quad \Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{x}-4 \mathrm{y}+5=0$

$(\mathrm{x}-4)^{2}+(\mathrm{y}-3)^{2}=0 \quad \Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}-8 \mathrm{x}-6 \mathrm{y}+25=0$

System of co-axial of circles equation is

$x^{2}+y^{2}-2 x-4 y+5+\lambda\left(x^{2}+y^{2}-8 x-6 y+25\right)=0$

equation (1) passes through origin

$\therefore 5+25 \lambda=0$

$\therefore \lambda=-1 / 5$

Substituting in (1) we get

$4\left(x^{2}+y^{2}\right)-2 x-14 y=0$

$2 x^{2}+2 y^{2}-x-7 y=0$



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