Radical Axis
The radical axis of two
circles is the locus of a point from which the tangent segments to the two
circles are of equal length.
Equation to the Radical
Axis
In general S - S' = 0
represents the equation of the Radical Axis to the two
circles i.e,2x(g - g ¢) + 2y(f - f ¢)
+ c - c¢ =0 where S º x2 + y2 + 2gx
+ 2fy + c=0 and
S' º x2 + y2 + 2g'x + 2f'y + c' =0.
·
If S = 0 and S' =
0 intersect in real and distinct points then S - S' = 0 is the equation of the
common chord of the two circles.
·
If S' = 0 and S =
0 touch each other, then S - S' = 0 is the equation of the common tangent to
the two circles at the point of contact.
Example 1
Prove
that the circle x2 + y2 – 6x – 4y + 9 = 0 bisects the
circumference of the circle x2 + y2 – 8x – 6y + 23 = 0.
Solution:
The given circles are
S1 º
x2 + y2 – 6x – 4y + 9 = 0 ………..(1)
.
and S2 º
x2 + y2 – 8x – 6y + 23 = 0. …….….(2).
Equation of the common
chord of circles (1) and (2) which is also the radical axis of the circles S1
and S2 is
S1 – S2
= 0 or, 2x + 2y – 14 = 0
or, x + y – 7 = 0 ………..(3).
Centre of the circle S2
is (4, 3). Clearly, line (3) passes through the point (4, 3) and hence line
(3) is the equation of the diameter of the circle (2). Hence circle (1) bisects
the circumference of circle (2).
Example 2
If two circles cut a third circle
orthogonaly, prove that their common chord will pass through the centre of the
third circle. .
Solution:
Let
us take the equation of the two circles as
x2 + y2 + 2l1x + a = 0 ….(1) .
x2 + y2 + 2l2x + a = 0 ….(2) .
We can select axes suitable (the line
of centres as x–axis and the point midway between the centres as origin) to get
the above form of equation. .
Let the third circle be x2
+ y2 + 2gx + 2fy + c = 0 ….(3) .
The circle (1) and (3) cut
orthogonally 2l1g = a + c ….(4) .
The circles (2) and (3) cut
orthogonally
2l2 g = a + c …(5)
From (4) and (5), 2g(l1 – l2) = 0 but l1 > l2
\
g = 0
Hence centre of the third circle (0,
–f)
The common chord of (1) and (2) has
the equation S1 – S2 = 0
i.e. (x2 + y2 +
2l1x
+ a) – (x2 + y2 + 2l2x + a) = 0 .
or 2(l1 – l2) x = 0 \
x = 0
\ (0, –f) satisfies the equation x =
0.
