Example 1.
Find the coordinates of
the points at which the circles x2 + y2 - 4x - 2y = 4 and
x2 + y2 - 12x - 8y = 12 touch each other. Find the
coordinates of the point of contact and equation of the common tangent at the
point of contact.
Solution
The given
circles are
S1
º
x2 + y2 – 4x – 2y – 4 = 0 and S2 º
x2 + y2 – 12x – 8y – 12 = 0
Equation of
the common tangent is S1 – S2 = 0 i.e. 8x + 6y + 8 = 0.
&⇒ 4x + 3y + 4 = 0 . . .
(1).
Let the
point of contact be (x1, y1)
Equation of
the tangent at (x1, y1) is ,
xx1
+ yy1 – 2(x + x1) – (y + y1) – 4 = 0
or (x1
– 2)x + (y1 – 1)y – 2x1 – y1 – 4 = 0 .
. . (2).
Comparing
(1) and (2), we get
&⇒ x1 = 4k + 2, y1
= 3k + 1, 2x1 + y1 + 4 = – 4k
&⇒ 2(4k + 2) + 3k + 1 + 4 = – 4k or
8k + 4 + 3k + 5 = – 4k
&⇒ 15k = –9
&⇒
k = 
&⇒
x1 = 4 ´ 
,
y1 = 3 ´ 
Point of
contact = 
Example 2.
A variable triangle has
two of its sides along the axes. Its 3rd side touches the circle x2
+ y2 - 2ax - 2ay + a2 = 0. Prove that the locus of the
circumcentre of the triangle is a2 - 2a(x + y) + 2xy = 0.
Solution
The given
circle has its centre (a, a) and radius a, so that the circle touches both
the axes. Let the side AB be 
&⇒
A º (p, o), B º (o, q)
Let R be the
midpoint of AB, so that
R º
=
(h, k)
&⇒ p = 2h, q = 2k
Since the
line touches
the circle,
|
So, 
&⇒
(ap + aq –
pq)2 = a2 (p2 +q2)
&⇒ a2(p+q)2+p2q2–2a(p+q)×pq
= a2(p2 + q2)
&⇒ 2a2 +pq – 2a (p + q) =
0
&⇒ a2 + 2hk – 2a (h + k) =
0
Locus of
(h, k) is
a2
+ 2xy – 2a (x + y) = 0
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|
Example 3.
Consider a curve ax2
+ 2hxy + by2 = 1, and a point P not lying on the curve. A line drawn
from the point P intersects the curve at points Q and R. If the product PQ.PR
is independent of the slope of the line, show that the curve is a circle.
Solution
Let P be the point (a,
b).
The equation of the line through P, with slope tanq,
is 
=
r
Any point on this line
is (a + r cosq, b + r sinq).
If this point lies on the curve, then .
a(a
+ r cosq)2 + 2h(a + r cosq)
(b
+ r sinq) + b(b + r sinq)2
= 1
This is a quadratic equation in r
whose values give the coordinates of Q and R, where PQ = r1 and PR =
r2.
&⇒
r1r2 = 
= 
which is independent of
q
if 
+
h2 = 0
&⇒ a = b, h = 0.
Hence the given curve
is a circle.
Example 4.
Find the equation of
the circle passing through (-4, 3) and touching the lines
x + y = 2 and x - y = 2.
Solution
|
The
lines are mutually perpendicular and passes through (2, 0). .
Due
to symmetry, obviously the centre of the required circle lies on x-axis. Let
it be (-a, 0).
So
equation of circle is
x2
+ y2 + 2ax + c = 0
It
passes through (-4, 3)
\ 16 + 9 – 8a + c =
0
&⇒ 8a – 25 = c
|
|
radius = 
But radius = length of
perpendicular from (-a, 0) on x + y = 2. .
This gives a = 10 ±
3
&⇒
c = 55 ± 24
circle is x2 + y2
+ 2(10 ± 3)x
+ 55 ± 24=
0
Example 5.
Find the
circumcentre of the triangle formed by x + y =0, x – y = 0 and lx + my =1. If l
and m vary such that l2 + m2 = 1, show that the locus of
its circumcentre is the curve (x2 – y2)2 = x2
+ y2 .
Solution
Solving the straight lines x – y = 0,
x + y = 0, lx + my = 1
we get the vertices as A(0, 0), B
and
if S(h, k) is the circumcentre then
SA2 = SB2 = SC2 and hence
h2 + k2
= 
=
or h + k = 
….(1)
or h – k = 
….(2)
if l, m vary such that l2
+ m2 = 1, then the locus of the circumcentre is got from
h2 + k2
= 
=
(h2 – k2)2
Hence locus of circumcentre is (x2
– y2)2 = x2 + y2 . .
