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
`11 and 12 Grade Math
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