Example
1.
A circle of
diameter 13 m with the centre O coinciding with the origin of co-ordinate axes,
has the diameter AB on the x-axis (x co-ordinates of B > 0). If the length
of the chord AC be 5m, find the equation of pair of lines BC, C having two possible
positions. .
Solution
Given AB = 13m, AC = 5m. Let OL ^AC,
then L is the mid-point of AC. .
Now AL =
and AO = 
From D ALO, cosq
=
\ slope of BC
=tan(90° + q) = – cotq
=
Again slope of BC¢
= tan(– 90°
– q) = cotq =
\
Equation of BC and BC¢ are 
or y
and 
or 
\their
joint equation is y2 -
or 
or 576y2 –100x2
+1300x –4225 = 0
100x2 – 576y2
–1300x + 4225 = 0 .
Example
2.
If S1
º
x2 + y2 +2g1x + 2f1y +c1
= 0 and S2 º x2 + y2 +2g2x
+ 2f2y +c2 = 0 are two circles with radii r1 and
r2 respectively, show that the points at which the circles subtend
equal angles lie on the circle 
.
Solution
If the angle subtended
by the circles S1 = 0 and S2 = 0 at point A are q
then
∠SAR
= ∠PAQ = q
\
∠C1AR
= ∠PAC2 = 
Let A(h, k). Then AR = 
,
AP = 
.
&⇒
&⇒
\
locus of (h, k) is 
i.e.
.
Example
3.
A chord of
the circle x2 + y2 - ax - by = 0, drawn from the point
(a, b), is divided by the x-axis in the ratio 2 : 1. Prove that a2 >
3b2. .
Solution
The given point A(a, b) lies on the given
circle. Any point on the circle is P
.
The x-axis cuts the chord AP in the ratio 2:1. Hence 
=
0
&⇒ sinq = -
&⇒ |sinq| = 
<
1
&⇒ 4b2 <
a2 + b2
&⇒
a2 > 3b2 .
Example
4.
Find the equation to
the straight lines joining the origin to the points in which the straight line
y = mx +c cuts the circle x2 + y2 =2ax + 2by. .
Hence find
the condition that these points may subtend a right angle at the origin. Find
also the condition that the straight line may touch the circle.
Solution
The straight line is
given as y =mx + c or
….(1)
The equation of the circle is given
as
x2 + y2 = 2ax +
2by or x2 + y2 –2ax – 2by =0 ….(2).
Making the second equation
homogeneous with the help of (1), we get the combined equation of the straight
lines joining the points of intersection of (1) with (2) to the origin, hence
the required equation is
x2 + y2 –2 ax
= 0
or cx2 + cy2
–2ax(y – mx) – 2by(y –mx) = 0
or x2 ( c+2am) + y2
(c –2b) –2xy(a –bm) = 0 . . . (3).
If the lines represented by this
equation are at right angles, the sum of the coefficients of x2 and
y2 should be zero. Hence the required condition is
c + 2am + c –2b = 0 or c = b – am.
If the line (1) touches the circle,
it isonly possible when the line cuts the circle in two coincident points.
Therefore the lines joining the origin to the two points will also coincide.
Hence the equation (3) must represent one line only. It is only possible when
it is a perfect square, so its discriminant must be zero. .
So [ 2 (a – bm)]2 –4(c
+2am) (c – 2b) = 0
Or a2 + b2m2
– 2abm –c2 +2bc –2amc + 4abm = 0
Or c2 – 2c (b – am) –a2
- b2m2 – 2abm = 0
Or c2 – 2c (b –am) – (a +
bm)2 =0
\
c = 
= b – am ±
Ö
[ ( b2 + a2 ) ( 1+m2 )] . .
Example
5.
DABC
has vertices A(a cosa, a sina), B( a cosb,
a sinb) and C(a cosg, a sing). Prove that
its orthocentre is given by .
H{a(cosa
+ cosb + cosg), a(sina
+ sinb + sing)}. .
Solution
The DABC is
inscribed in circle x2 + y2 = a2
so circumcentre is origin O
centroid is 
Centroid divide line going
circumcentre and orthocentre (h, k) in ratio 1 : 2. .
\
&⇒
h = a(cosa + cosb + cosg)
Similarly k = a(sina
+ sinb + sing).
