Example 1.
A rectangle
ABCD is inscribed in a circle with a diameter lying along the line
3y = x + 10. If A and B are the points (-6, 7) and (4, 7) respectively, find
the area of the rectangle.
Solution
|
Let O (h, k) be the centre of circle.
OA2 = OB2.
&⇒
(h + 6)2 + (k – 7)2 =(h–4)2 +(k – 7)2
&⇒
h2 +36 + 12h = h2 + 16 – 8h
&⇒
20h = –20
&⇒ h = –1
(h,
k) also lies on the line 3y = x + 10
&⇒
3k = h + 10
&⇒ 3k = 9
&⇒
k = 3
&⇒ O º (-1, 3)
|
|
Now AB =
,
OB = 
=
and
BD = 
.
In DADB, BD2
= AD2 + AB2 i.e. 164 = AD2 + 100.
&⇒
AD2 = 64
&⇒ AD = 8
Area of
Rectangle ABCD = 10 ´ 8 = 80 sq. units.
Example 2.
AB is the
diameter of a circle, CD is a chord parallel to AB and 2 CD = AB. The tangent
at B meets the line AC produced at E. Prove that AE = 2AB.
Solution
|
Let O, the centre of circle, be taken as the
origin. A
(-a, 0) and B (a, 0) a is radius of the .
circle. In DOCF,.
OC2=CF2+OF2
(Since CD= AB
and CD|| AB)
&⇒
a2 =  +
OF2
&⇒ OF2 = 
&⇒
OF = ± 
&⇒
C 
Equation
of the line AE where A (-a, 0),  is
y – 0 = 
|
|
y = Ö3
(x + a) … (1)
Equation of line BE
[which is tangent at (a, 0) is
x = a …
(2)
Point of intersection
of (1) and (2) is E i.e. E (a, 2Ö3a).
AE = 
=
4a = 2(2a) = 2AB
Hence AE =
2AB
Example 3.
Find the
locus of centres of the circle which touches the two circle x2 + y2
= a2 and x2 + y2 = 4ax externally. .
Solution
Let (h, k) be the centre of the
circle and r be its radius then 
=
r + a
and 
Eliminating r, 
\
locus of (h, k) is 
Squaring (x – 2a)2 + k2
= a2 + x2 + y2 + 2a
&⇒
- 4ax + 3a2 = 2a
Squaring we
get 12 x2 – 4y2 – 24ax + 9a2 = 0.
Example 4.
If 
i
= 1, 2, 3, 4 mi > 0 are 4 distinct points on a circle, then show
that
m1 .m2 .m3 .m4 = 1.
Solution
where i = 1, 2, 3, 4 are four points
lying on a circle.
Let
the equation of the circle be x2 + y2 + 2gx + 2fy + c = 0
is lying on the circle
&⇒ mi2 + 
+
2gmi + 
=
0
&⇒ mi4 + 2gmi3
+ 2fmi + cmi2 + 1 = 0
If m1, m2, m3,
m4 are its roots then
&⇒ m1 ×
m2 × m3 ×
m4 = 1
Example 5.
If al2
– bm2 + 2dl + 1 = 0 a, b, d are fixed real numbers such that a + b =
d2, then prove that the line lx + my + 1 = 0 touches a fixed circle.
Find its equation.
Solution
Let the equation of the circle be x2
+ y2 + 2gx + 2fy + c = 0
If lx + my + 1 = 0 is a tangent to
the circle
Then 
&⇒
(g2 + f2 – c) (l2 + m2) = (lg + mf
– 1)2
&⇒
(f2 – c) l2 + (g2 – c)m2 + 2l (g –
mgf) + 2mf – 1 = 0
Comparing with the given condition al2
– bm2 + 2dl + 1 = 0,
– (f2 – c) = a, – (g2
– c) = – b, -g (1 – mf) = d, f = 0
&⇒
c = a, g = – d, g2 – c = b, d2 – a = b
&⇒
a + b = d2 which is the required condition.
Hence the
fixed circle is x2+ y2 – 2dx + a = 0.
