Find the radical axis of two circles and radical centre of three circles

Question

Find the radical axis of the circles $S_1: x^2 + y^2 - 4x - 6y + 9 = 0$ and $S_2: x^2 + y^2 - 2x - 4y + 1 = 0$ . Also find the radical centre of $S_1$ , $S_2$ , and $S_3: x^2 + y^2 + 6x - 2y - 3 = 0$ .

(JEE Advanced 2022, similar pattern)

Solution — Step by Step

The radical axis of two circles $S_1 = 0$ and $S_2 = 0$ is simply $S_1 - S_2 = 0$ .

(x^2 + y^2 - 4x - 6y + 9) - (x^2 + y^2 - 2x - 4y + 1) = 0

-4x + 2x - 6y + 4y + 9 - 1 = 0

-2x - 2y + 8 = 0

\boxed{x + y = 4}

S_2 - S_3 = (x^2 + y^2 - 2x - 4y + 1) - (x^2 + y^2 + 6x - 2y - 3) = 0

-8x - 2y + 4 = 0

4x + y = 2

The radical centre is the intersection of any two radical axes. Solve:

x + y = 4 \quad \cdots (1)

4x + y = 2 \quad \cdots (2)

Subtract (1) from (2): $3x = -2 \implies x = -\frac{2}{3}$

From (1): $y = 4 - (-\frac{2}{3}) = \frac{14}{3}$

\boxed{\text{Radical centre} = \left(-\frac{2}{3}, \frac{14}{3}\right)}

Why This Works

The radical axis of two circles is the locus of points having equal power with respect to both circles. The power of a point $(h, k)$ with respect to circle $S = 0$ is $S_1 = h^2 + k^2 + 2gh + 2fk + c$ .

Setting $S_1(\text{power w.r.t. first}) = S_2(\text{power w.r.t. second})$ gives $S_1 - S_2 = 0$ , which is always a straight line (the $x^2$ and $y^2$ terms cancel).

The radical centre is the unique point with equal power with respect to all three circles. Geometrically, it’s where the three radical axes (taken pairwise) meet. All three radical axes are concurrent — this is a beautiful theorem in circle geometry.

If the radical centre lies outside all three circles, a single circle centred at the radical centre can be drawn that is orthogonal to all three circles.

Alternative Method — Using power of a point directly

For any point $P(h, k)$ on the radical axis of two circles with centres $C_1$ , $C_2$ and radii $r_1$ , $r_2$ :

PC_1^2 - r_1^2 = PC_2^2 - r_2^2

This gives the same linear equation as $S_1 - S_2 = 0$ .

In JEE, the radical axis has a key property: if two circles intersect, their radical axis is the common chord. If they touch, it’s the common tangent at the point of tangency. If they don’t intersect, it’s still a line (between the circles), but no point of the line lies on either circle. Recognising which case applies can shortcut the problem.

Common Mistake

When computing $S_1 - S_2$ , students sometimes forget to subtract ALL terms and miss the constant term. Write out the subtraction carefully, term by term. Also, some students try to find the radical axis by finding intersection points of the circles — this only works when the circles actually intersect. The $S_1 - S_2 = 0$ method works regardless of whether the circles intersect, are tangent, or are separate.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using power of a point directly

Common Mistake

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