Family of Circles
- If S º x2 + y2 + 2gx
+ 2fy + c = 0 and S¢ º x2
+ y2 + 2g¢x + 2f¢y + c¢
= 0 are two intersecting circles, then S + lS¢
= 0, l > -1, is the equation of the family of
circles passing through the points of intersection of S = 0 and S¢
= 0.
- If S º x2
+ y2 + 2gx + 2fy + c = 0 is a circle which is intersected by the
straight line
m
º
ax + by + c = 0 at two real and distinct points, then S + lm = 0 is the equation of the family of
circles passing through the points of intersection of S = 0 and m=0.
Ifm = 0 touches S = 0 at P, then S + lm = 0 is the equation of the family of
circles, each touching m = 0 at P. - The equation of a family of circles
passing through two given points (x1, y1) and
(x2, y2) can be written in the form - (x - x1)(x - x2) + (y - y1)(y
- y2) +
where l
is a parameter. - The equation of the family of circles
which touch the line y - y1 = m(x - x1) at
(x1, y1) for any value of m is (x - x1)2 +
(y - y1)2 + l[(y - y1) -m(x - x1)]
= 0. If m is infinite, the equation is (x - x1)2 + (y - y1)2
+ l(x - x1) = 0. - The two circles are said to intersect
orthogonally if the angle of intersection of the circles i.e., the angle
between their tangents at the point of intersection is 90° .
The condition for the
two circles S = 0 and S1= 0 to cut each other orthogonally is 2gg1
+ 2ff1 = c + c1 . .
Example 1
Find
the equation of the circle described on the common chord of the circles x2
+ y2 - 4x - 5 = 0 and x2 + y2 + 8y + 7 = 0 as
diameter.
Solution:
Equation of the common chord is S1
- S2 = 0
&⇒ x + 2y + 3 = 0
Equation of the circle
through the two circles is S1 + lS2 =
0
&⇒
x2 + y2 - 
x + 
+ 
= 0
Its centre 
lies on x + 2y + 3 = 0
&⇒
+ 3 = 0
&⇒
2 - 8l + 3 + 3l = 0
&⇒
l
= 1. Hence the required circle is x2 + y2 - 2x + 4y + 1
= 0.
Example 2
Show
that the circle passing through the origin and cutting the circles x2
+ y2 - 2a1x - 2b1y + c1 = 0 and x2
+ y2 - 2a2x - 2b2y + c2 = 0
orthogonally is 
= 0 .
Solution:
Let the equation of the circle
passing through the origin be
x2 + y2 + 2gx +
2fy = 0, . . . . (1).
It cuts the given two circles
orthogonally
&⇒ -2ga1 - 2fb1 =
c1
&⇒
c1 + 2ga1 + 2fb1 = 0, .
. . . . (2).
and -2ga2 - 2fb2
= c2
&⇒
c2 + 2ga2 + 2fb2 = 0. .
. . . . . (3).
Eliminating 2f and 2g from (1), (2)
and (3), we get
=
0 .
Example 3
Tangents PQ and PR
are drawn to the circle x2 + y2 = a2 from the
point P(x1, y1). Prove that equation of the circum circle
of DPQR is
x2 + y2 –xx1 – yy1= 0.
Solution:
QR is the chord of contact of the
tangents to the circle
x2 + y2 – a2
= 0 ….(1).
equation of QR is xx1 + yy1
– a2 ….(2) .
The circumcircle of DPQR
is a circle passing through the intersection of the circle (1) and the line (2)
and the point P(x1, y1). .
Circle through the intersection of
(1) and (2) is
x2 + y2 – a2
+ l (xx1 + yy1 –a2) = 0 …(3)
it will pass through (x1,
y1) if
x12 + y12
– a2 + l (x12 + y12
–a2) = 0
l
= –1 (since x12 + y12 >
a2)
Hence equation of circle is (x2
+ y2 – a2) – (xx1 + yy1 –a2)
= 0
Or x2 + y2 –
xx1 – yy1 = 0 . .
