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 + m^{2}) (x^{2} +y^{2})
–2a(x+ my) = 0. .
Solution
(a, 0) is centre of given circle
whose equation is (x – a)^{2} + y^{2} = a^{2}
i.e. x^{2} + y^{2}
– 2ax = 0 .
Equation of circle with OP (y = mx)
as diameter is
x^{2} + y^{2} – 2ax
+ l(y – mx) = 0
whose centre is


This centre must lie on y = mx
\
&⇒
l
= 
\
centre is x^{2} + y^{2} – 2ax  (y
– mx) = 0
i.e. (1 + m^{2}) (x^{2}
+ y^{2}) – 2a(1 + m^{2})x – 2amy + 2am^{2}x = 0 .
i.e. (1 + m^{2})(x^{2}
+ y^{2}) – 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
xaxis and a circle is drawn with centre at P which touches C externally and
meets the xaxis in points M and N. Find the coordinates of a point Q on the
yaxis 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} = d^{2}
+ h^{2}
Also,
slope of the line
Slope
of line NQ = .
Also q = q_{2}  q_{1 }
Put
(r + s)^{2} = d^{2} + h^{2} to eliminate h
&⇒


If we take k^{2} = d^{2}
 r^{2}, then
Hence for k^{2} = d^{2}
 r^{2}, ∠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} = b^{2}, (x – b)^{2} +
(y – a)^{2} = a^{2} and (x – a –b – c)^{2} + y^{2}
= ab + c^{2} is
x^{2} + y^{2} – 2x (a + b) – y (a + b) + a^{2} + 3ab +
b^{2} = 0.
Solution
The three circles are x^{2}
+ y^{2} – 2ax – 2by + a^{2} = 0
x^{2} + y^{2}
– 2bx – 2ay + b^{2} = 0
x^{2} + y^{2}
– 2(a+ b + c)x + a^{2} + b^{2} + ab + 2bc + 2ac = 0
Let x^{2} + y^{2}
+ 2gx + 2fy + k = 0
Cut them orthogonally,
then
2g(a) + 2f(b) = k + a^{2}
….(1).
2g(b) + 2f(a) = k + b^{2}
….(2).
2g(a – b  c) + 2f.0 =
k + a^{2} + b^{2} + ab + 2c(a + b) ….(3).
Eliminating f from (1)
and (2), we get
 2g(a + b) = k + a^{2}
+ ab + b^{2}
Substituting from (3),
 2gc = 2c(a + b)
&⇒  (a + b) = g
\
k = 2(a + b)^{2} – a^{2} – ab – b^{2} = 2(a^{2}
+ 2ab + b^{2}) – a^{2} – ab – b^{2}
= a^{2} + b^{2}
+ 3ab
\
(1)
&⇒ f = 
so the equation of
required circle is
x^{2} + y^{2}
– 2(a + b)x – (a + b)y + a^{2} + b^{2} + 3ab = 0
Example
4.
Is (3, 2)
an interior point or an exterior point of the circle x^{2} + y^{2}
– 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
3^{2} + 2^{2}
– 6 + 4 > 0 so (3, 2) is exterior point of the circle
x^{2} + y^{2}
– 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 (acosq_{I}, a sinq_{i}); i = 1, 2, 3, 4 represents the
points A, B, C and D respectively. .
The orthocentre of DABC
is (a(cosq_{1} + cosq_{2} + cosq_{3}), a (sinq_{1} + sinq_{2} + sinq_{3}))
This will lies on the circle
[x  a(cosq_{1} + cosq_{2} + cosq_{3})]^{2} + [y – a(sinq_{1} + sinq_{2} + sinq_{3})]^{2} = a^{2}
because (a cosq_{4})^{2} + (a sinq_{4})^{2} = a^{2} is
true
Similarly
the orthocentre of DBCD, DACD, DABD
all lie on circle (1).