Example
1.
Let y = mx
be the equation of a chord of a circle whose radius is ‘a’, the origin of
coordinates being one extremity of the chord, and the axis of x being a
diameter of the circle. Prove that the equation of the circle of which this
chord is a diameter is (1 + m2) (x2 +y2)
–2a(x+ my) = 0. .
Solution
|
(a, 0) is centre of given circle
whose equation is (x – a)2 + y2 = a2
i.e. x2 + y2
– 2ax = 0 .
Equation of circle with OP (y = mx)
as diameter is
x2 + y2 – 2ax
+ l(y – mx) = 0
whose centre is 
|
|
This centre must lie on y = mx
\
&⇒
l
= - 
\
centre is x2 + y2 – 2ax - (y
– mx) = 0
i.e. (1 + m2) (x2
+ y2) – 2a(1 + m2)x – 2amy + 2am2x = 0 .
i.e. (1 + m2)(x2
+ y2) – 2ax – 2amy = 0.
Example
2.
C is a
circle with centre (0, d) and radius r (r < d). A point P is chosen on the
x-axis and a circle is drawn with centre at P which touches C externally and
meets the x-axis in points M and N. Find the coordinates of a point Q on the
y-axis such that ∠MQN is constant for any choice of the
point P.
Solution
Suppose P º
(h, 0) and Q º (0, k) and the radius is s.
&⇒
M º (h - s, 0) and N º (h + s , 0). .
Then (r + s)2 = d2
+ h2
Also,
slope of the line 
|
Slope
of line NQ =  .
Also q = q2 - q1 
Put
(r + s)2 = d2 + h2 to eliminate h
&⇒ 
|
|
If we take k2 = d2
- r2, then 
Hence for k2 = d2
- r2, ∠MQN is constant for all points P.
Example
3.
Show that
the equation to the circle cutting orthogonally the circles
(x –a)2 + (y – b)2 = b2, (x – b)2 +
(y – a)2 = a2 and (x – a –b – c)2 + y2
= ab + c2 is
x2 + y2 – 2x (a + b) – y (a + b) + a2 + 3ab +
b2 = 0.
Solution
The three circles are x2
+ y2 – 2ax – 2by + a2 = 0
x2 + y2
– 2bx – 2ay + b2 = 0
x2 + y2
– 2(a+ b + c)x + a2 + b2 + ab + 2bc + 2ac = 0
Let x2 + y2
+ 2gx + 2fy + k = 0
Cut them orthogonally,
then
2g(-a) + 2f(-b) = k + a2
….(1).
2g(-b) + 2f(-a) = k + b2
….(2).
2g(-a – b - c) + 2f.0 =
k + a2 + b2 + ab + 2c(a + b) ….(3).
Eliminating f from (1)
and (2), we get
- 2g(a + b) = k + a2
+ ab + b2
Substituting from (3),
- 2gc = 2c(a + b)
&⇒ - (a + b) = g
\
k = 2(a + b)2 – a2 – ab – b2 = 2(a2
+ 2ab + b2) – a2 – ab – b2
= a2 + b2
+ 3ab
\
(1)
&⇒ f = - 
so the equation of
required circle is
x2 + y2
– 2(a + b)x – (a + b)y + a2 + b2 + 3ab = 0
Example
4.
Is (3, 2)
an interior point or an exterior point of the circle x2 + y2
– 2x + y = 0. If interior, find the equation of the circle centred on it
and of maximum area contained in the given circle. If exterior find the
equation of the circle of minimum radius centred on it containing the given
circle. .
Solution
32 + 22
– 6 + 4 > 0 so (3, 2) is exterior point of the circle
x2 + y2
– 2x + y = 0 …(1)
The radius of required
circle
= distance between
centres (3, 2) and (1, -1/2) + radius of circle (1)
= 
=
\
Equation of required circle is (x – 3)2 + ( y – 2)2 = 
Example
5.
If ABCD is a
cyclic quadrilateral, prove that the orthocentre of the triangle ABC, BCD, CDA
and DAB lies on a circle.
Solution
Let (acosqI, a sinqi); i = 1, 2, 3, 4 represents the
points A, B, C and D respectively. .
The orthocentre of DABC
is (a(cosq1 + cosq2 + cosq3), a (sinq1 + sinq2 + sinq3))
This will lies on the circle
[x - a(cosq1 + cosq2 + cosq3)]2 + [y – a(sinq1 + sinq2 + sinq3)]2 = a2
because (a cosq4)2 + (a sinq4)2 = a2 is
true
Similarly
the orthocentre of DBCD, DACD, DABD
all lie on circle (1).
