We will learn how to find the equation of the common chord of two circles.

Let us assume that the equations of the two given intersecting circles be x\(^{2}\) + y\(^{2}\) + 2g\(_{1}\)x + 2f\(_{1}\)y + c\(_{1}\) = 0 ……………..(i) and x\(^{2}\) + y\(^{2}\) + 2g\(_{2}\)x + 2f\(_{2}\)y + c\(_{2}\) = 0 ……………..(ii), intersect at P (x\(_{1}\), y\(_{1}\)) and Q (x\(_{2}\), y\(_{2}\)).

Now we need to find the equation of the common chord PQ of the given circles.

Now we observe from the above figure that the point P (x\(_{1}\), y\(_{1}\)) lies on both the given equations.

Therefore, we get,

x\(_{1}\)\(^{2}\) + y\(_{1}\)\(^{2}\) + 2g\(_{1}\)x\(_{1}\) + 2f\(_{1}\)y\(_{1}\) + c\(_{1}\) = 0 ……………..(iii)

x\(_{1}\)\(^{2}\) + y\(_{1}\)\(^{2}\) + 2g\(_{2}\)x\(_{1}\) + 2f\(_{2}\)y\(_{1}\) + c\(_{2}\) = 0 ……………..(iv)

Now subtracting the equation (4) from equation (3) we get,

2(g\(_{1}\) - g\(_{2}\))x\(_{1}\) + 2 (f\(_{1}\) - f\(_{2}\))y\(_{1}\) + C\(_{1}\) - C\(_{2}\) = 0 ……………..(v)

Again, we observe from the above figure that the point Q (x2, y2) lies on both the given equations. Therefore, we get,

x\(_{2}\)\(^{2}\) + y\(_{2}\)\(^{2}\) + 2g\(_{1}\)x\(_{2}\) + 2f\(_{1}\)y\(_{2}\) + c\(_{1}\) = 0 ……………..(vi)

x\(_{2}\)\(^{2}\) + y\(_{2}\)\(^{2}\) + 2g\(_{2}\)x\(_{2}\) + 2f\(_{2}\)y\(_{2}\) + c\(_{2}\) = 0 ……………..(vii)

Now subtracting the equation (b) from equation (a) we get,

2(g\(_{1}\) - g\(_{2}\))x\(_{2}\) + 2 (f\(_{1}\) - f\(_{2}\))y\(_{2}\) + C\(_{1}\) - C\(_{2}\) = 0 ……………..(viii)

From conditions (v) and (viii) it is evident that the points P (x\(_{1}\), y\(_{1}\)) and Q (x\(_{2}\), y\(_{2}\)) lie on 2(g\(_{1}\) - g\(_{2}\))x + 2 (f\(_{1}\) - f\(_{2}\))y + C\(_{1}\) - C\(_{2}\) = 0, which is a linear equation in x and y.

It represents the equation of the common chord PQ of the given two intersecting circles.

**Note:** While finding the equation of the common chord
of two given intersecting circles first we need to express each equation to its
general form i.e., x\(^{2}\) + y\(^{2}\) + 2gx + 2fy + c = 0 then subtract
one equation of the circle from the other equation of the circle.

Solve example to find the equation of the common chord of two given circles:

**1.** Determine the equation of the
common chord of the two intersecting circles x\(^{2}\) + y\(^{2}\) - 4x
- 2y - 31 = 0 and 2x\(^{2}\) + 2y\(^{2}\) - 6x + 8y - 35 = 0 and prove
that the common chord is perpendicular to the line joining the centers of the
two circles.

**Solution:**

The given two intersecting circles are

x\(^{2}\) + y\(^{2}\) - 4x - 2y - 31 = 0 ……………..(i) and

2x\(^{2}\) + 2y\(^{2}\) - 6x + 8y - 35 = 0

⇒ x\(^{2}\) + y\(^{2}\) - 3x + 4y - \(\frac{35}{2}\) ……………..(ii)

Now, to find the equation of the common chord of two intersecting circles we will subtract the equation (ii) from the equation (i).

Therefore, the equation of the common chord is

x\(^{2}\) + y\(^{2}\) - 4x - 2y - 31 - (x\(^{2}\) + y\(^{2}\) - 3x + 4y - \(\frac{35}{2}\)) = 0

⇒ - x - 6y - \(\frac{27}{2}\) = 0

⇒ 2x + 12y +
27 = 0, which is the required equation.

The slope of the common chord 2x + 12y + 27 = 0 is (m\(_{1}\)) = -\(\frac{1}{6}\).

Centre of the circle x\(^{2}\) + y\(^{2}\) - 4x - 2y - 31 = 0 is (2, 1).

Centre of the circle 2x\(^{2}\) + 2y\(^{2}\) - 6x + 8y - 35 = 0 is (\(\frac{3}{2}\), -2).

The slope of the line joining the centres of the circles (1) and (2) is (m\(_{2}\)) = \(\frac{-2 - 1}{\frac{3}{2} - 2}\) = 6

Now m\(_{1}\) ∙ m\(_{2}\) = -\(\frac{1}{6}\) ∙ 6 = - 1

Therefore, we see that the slope of the common chord and slope of the line joining the centres of the circles (1) and (2) are negative reciprocals of each other i.e., m\(_{1}\) = -\(\frac{1}{m_{2}}\) i.e., m\(_{1}\) ∙ m\(_{2}\) = -1.

Therefore, the common
chord of the given circles is perpendicular to the line joining the centers of the
two circles. *Proved*

**●** **The Circle**

**Definition of Circle****Equation of a Circle****General Form of the Equation of a Circle****General Equation of Second Degree Represents a Circle****Centre of the Circle Coincides with the Origin****Circle Passes through the Origin****Circle Touches x-axis****Circle Touches y-axis****Circle Touches both x-axis and y-axis****Centre of the Circle on x-axis****Centre of the Circle on y-axis****Circle Passes through the Origin and Centre Lies on x-axis****Circle Passes through the Origin and Centre Lies on y-axis****Equation of a Circle when Line Segment Joining Two Given Points is a Diameter****Equations of Concentric Circles****Circle Passing Through Three Given Points****Circle Through the Intersection of Two Circles****Equation of the Common Chord of Two Circles****Position of a Point with Respect to a Circle****Intercepts on the Axes made by a Circle****Circle Formulae****Problems on Circle**

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