Radical Axis: Definition, Equation, Formula, Examples

Radical Axis

The radical axis of two circles is the locus of a point which moves in a plane in such a way that the lengths of the tangents drawn from it to the two circles are same.

Consider the two circle:

$\begin{array}{rl}& S_1:x^2+y^2+2g_{1}x+2f_{1}y+c_1=0\\ & S_2:x^2+y^2+2g_{2}x+2f_{2}y+c_2=0\end{array}$

Let point $\mathrm{P}({\mathrm{x}}_1,{\mathrm{y}}_1)$ such that

$|\mathrm{PA}|=|\mathrm{PB}|$

Length of the tangent from a point $\mathrm{P}({\mathrm{x}}_1,{\mathrm{y}}_1)$ to circle S is $\sqrt{S_1}$.

$\Rightarrow \sqrt{x_1^2+y_1^2+2g_1{x}_1+2f_1{y}_1+c_1}=\sqrt{x_1^2+y_1^2+2g_2{x}_1+2f_2{y}_1+c_2}$

On squaring both sides, we get

$\begin{array}{rl}& \Rightarrow {\mathrm{x}}_1^2+{\mathrm{y}}_1^2+2{\mathrm{g}}_1{\mathrm{x}}_1+2{\mathrm{f}}_1{\mathrm{y}}_1+{\mathrm{c}}_1={\mathrm{x}}_1^2+{\mathrm{y}}_1^2+2g_2{\mathrm{x}}_1+2{\mathrm{f}}_2{\mathrm{y}}_1+{\mathrm{c}}_2\\ & \Rightarrow 2{\mathrm{x}}_1({\mathrm{g}}_1-{\mathrm{g}}_2)+2{\mathrm{y}}_1({\mathrm{f}}_1-{\mathrm{f}}_2)+{\mathrm{c}}_1-{\mathrm{c}}_2=0\end{array}$
This is the required equation of radical axis.

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Solved Examples Based on Radical Axis:

Example 1: The equation of a circle which passes through $(2\mathrm{a},0)$ and whose radical axis in relation to the circle ${\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2$ is $\mathrm{x}=\mathrm{a}/2$, is
1) $x^2+y^2-ax=0$
2) $x^2+y^2+2ax=0$
3) $x^2+y^2-2ax=0$
4) $x^2+y^2-2ax=0$

Solution

The required circle is

${\mathrm{x}}^2+{\mathrm{y}}^2-{\mathrm{a}}^2+\lambda (x-\frac{a}{2})=0(\text{using}\mathrm{S}+\lambda \mathrm{L}=0)$

This passes through $(2\mathrm{a},0)$

$4a^2-a^2+\left(\frac{3a}{2}\right)\lambda =0\Rightarrow \lambda =-2a$

Hence, the required circle is

$\begin{array}{rl}& x^2+y^2-a^2-2a(x-\frac{a}{2})=0\\ & x^2+y^2-a^2-2ax+a^2=0\Rightarrow x^2+y^2-2ax=0\end{array}$

Hence, the answer is the option 3.

Example 2: The gradient of the radical axis of the circles $x^2+y^2-3x-4y+5=0$ and $3x^2+3y^2-7x+8y+11=0$ is:
1) $\frac{1}{3}$
2) $-\frac{1}{10}$
3) $-\frac{1}{2}$
4) $-\frac{2}{3}$

Solution

.The equation of the radical axis is:

$S_1-S_2=0S_1\equiv x^2+y^2-3x-4y+5=0,S_2\equiv x^2+y^2-\frac{7}{3}x+\frac{8y}{3}+\frac{11}{3}=0\gamma$

$\therefore$ Radical axis is $-2x-20y+4=0$.
Hence, the gradient of the radical axis $=-\frac{1}{10}$.
Hence, the answer is the option (2).

Example 3: The equation of the circle, which passes through the point (2a, 0 ) and whose radical axis is $\mathrm{x}=\frac{\mathrm{a}}{2}$ with respect to the circle ${\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2$, will be:
1) $x^2+y^2-2ax=0$
2) $x^2+y^2+2ax=0$
3) $x^2+y^2+2ay=0$
4) $x^2+y^2-2ay=0$

Solution

The equation of the radical axis is $\mathrm{x}=\frac{\mathrm{a}}{2}\Rightarrow 2\mathrm{x}-\mathrm{a}=0$
The equation of the required circle is $x^2+y^2-a^2+\lambda (2x-a)=0$
It passes through the point $(2\mathrm{a},0),\therefore 4{\mathrm{a}}^2-{\mathrm{a}}^2+\lambda (4\mathrm{a}-\mathrm{a})=0\Rightarrow \lambda =-\mathrm{a}$
$\therefore$ Equation of circle is $x^2+y^2-a^2-2ax+a^2=0\Rightarrow x^2+y^2-2ax=0$
Hence, the answer is the option (1).

Example 4: The square of the tangent that can be drawn from any pt. on one circle to another circle is k times the product of the perpendicular distance of the point from the radical axis of the two circles, and the distance between their centres, where $\mathrm{k}=$
1) 1
2) 3
3) 4
4) 2

Solution

We have to prove that ${\mathrm{PA}}^2=2\mathrm{PM}\cdot {\mathrm{C}}_1{\mathrm{C}}_2$

$\begin{array}{rl}& P\equiv (a\cos \theta ,a\sin \theta )\\ & PA=\sqrt{a^2-b^2+h^2-2ah\cos \theta }\end{array}$

Radical axis: $-2hx+h^2-a^2+b^2=0$ or $x=\frac{h^2-a^2+b^2}{2h}$

$$\begin{array}{rl}& PM=\frac{h^2-a^2+b^2}{2h}-a\cos \theta =\frac{h^2-a^2+b^2-2ah\cos \theta }{2h}\\ & PM\cdot C_1{C}_2=\frac{h^2-a^2+b^2-2ah\cos \theta }{2h}(\text{where}{C}_1{C}_2=h)=P{A}^2/2\end{array}$$

Example 5: The radical axis of the two distinct circles $x^2+y^2+2gx+2fy+c=0$ and $2x^2+2y^2+4x+y+2c=0$ touches the circle $x^2+y^2-4x-4y+4=0$. Then the centre of the circle $x^2+y^2+2gx+2fy+c=$ 0 can be
1) $(-1,-2)$
2) $(-2,\frac{1}{2})$
3) $(-2,\frac{1}{4})$
4) $(-1,\frac{1}{4})$

Solution

The radical axis of given circles is

$(2g-2)x+(2f-\frac{1}{2})y=0$

This line is tangent to the circle $x^2+y^2-4x-4y+4=0$
$\Rightarrow$ either $g=1$ or $f=\frac{1}{4}$
$\Rightarrow$ either $(g=1$ and $f\ne \frac{1}{4})$ or $(g\ne 1$ and $f=\frac{1}{4})$
Hence, the answer is the option (1).

Summary

The radical axis of two circles is a fundamental concept in circle geometry, providing a locus of points with equal power to both circles. By understanding its properties, mathematical formulation, and applications, one can gain deeper insights into the interactions between circles and their geometric relationships.