Example
1.
Find the locus of the
point of intersection of tangents to the circle x = acos q,
y = asinq at the point whose parametric angles differ by (i) p/3
(ii) p/2.
Solution
Let A and B be two
points on circle whose parametric angle differs by q
|
then PA = 
tan .
. . . . (1)
(i)
when q = p/3 then
eq.
(1) becomes h2 +k2 –a2 = a2/3.
locus of point
of intersection P is x2+y2 =
(ii)
when q = p/2 eq.(1) becomes .
h2
+k2 –a2 = a2
|
|
locus of point of
intersection P is x2 +y2 = 2a2
Example
2.
A point moves so that
the sum of the squares of its distances from n fixed point is given. Prove that
its locus is a circle.
Solution
Let (h, k) be the
moving point and (ai, bi); i = 1, 2, 3, …, n be the fixed
points. Then given 
(constant)
&⇒
&⇒
n ( h2 + k2) – 2h Sai – 2kSbi
+ S(ai2 + bi2)
- l = 0
\
locus of (h, k) is x2 + y2 - 2a
x - 2b y + g = 0 which is a circle
where a
= 
,
b
= 
,
g
= 
Example
3.
Lines 5x + 12y - 10 =
0 and 5x - 12y - 40 = 0 touch circle C1 of diameter 6. If centre of
C1 lies in the first quadrant, find the equation of circle C2,
which is concentric with circle C1 and cuts an intercept of length 8
on these lines.
Solution
|
Since the
lines 5x +12y – 10 = 0 and 5x–12y– 40 =0 touch the circle, its centre lies on
one of the angle bisectors of the given lines.
i.e 
5x + 12y –
10 = ±(5x – 12y – 40)
&⇒ 24y = –30 or 10x = 50
&⇒
 or
x = 5
But the
centre lies in first quadrant
&⇒ x = 5
Since C1
touches 5x + 12y – 10 = 0,
|
|
&⇒ 25 + 12h – 10
= ± 39
&⇒ 12h = ± 39 – 15
&⇒ 12h = 24
&⇒
h = 2 also h = 
(not
possible)
centre of C1
= (5, 2)
Let r be
radius of C2
&⇒ r2 = OP2 + AP2
= 32 + 42 = 52 or r = 5
Equation of
C2 is (x – 5)2 + (y – 2)2 = 52
&⇒
x2 + y2 – 10x – 4y + 4 = 0
Example 4.
A and B are two points
on the x-axis and the y-axis respectively. Two circles are drawn passing
through the origin and having centres at A and B respectively. Show that AB
bisects the common chord of these circles.
Solution
Let A(a, 0) and B(0,
b). Then the two circles are .
x2 + y2
– 2ax = 0 …(1)
and x2 + y2
– 2by = 0
Common chord is ax – by
= 0 ….(2) .
Equation of circle
through (1) and (2) is x2 + y2 – 2ax + l(ax
– by) = 0
i.e. x2 + y2
+ ax(l - 2) - l by = 0 .
This will be circle
with common chord (2) as diameter if centre 
lies
on (2) i.e. 
&⇒
l
= 
Mid-point of common
chord is
i.e. 
This lie on line AB (
equation 
)
if
which is true
so AB passes through
mid-point of common chord.
Example
5.
Prove that the locus of
the centre of a circle which touches the circle
x2 + y2 – 6x – 6y + 14 = 0 externally and touches the
y-axis is given by
y2 – 10x –6y + 14 = 0. .
Solution
Centre of given circle is (3, 3) and
radius = 2. .
Let r = radius and (h, k) be the
centre of other circle. .
r = length of perpendicular from (h,
k) on x = 0 (y-axis)
\
r = h
Equation is x2 + y2
– 2hx – 2ky – h = 0
Again r + 2 = distance between the
centres
&⇒
h + 2 = 
&⇒
(h + 2)2 = (h – 3)2 + (k - 3)2
&⇒
k2 – 6k – 10h + 14 = 0
\
locus of centre (h, k) is y2 - 6k – 10x + 14 = 0.
