Example
1.
Find the
equation of the tangent at A to the circle ( x - a)2 + (y - b)2
= r2, where the radius through A makes an angle a
with the x - axis.
Solution
The parametric
equations of the circle
(x–a)2 + (y
– b)2 = r2 are x = a + rcosa and y = b +
rsina
Then 
and
&⇒
Now the equation of the
tangent at (a + rcosa, b + rsina)
to the given circle is
y – (b + rsina) = – cot a
[x – (a + rcosa)]
or, (y – b) – rsina
= – (x – a) cota + rcos a
cot a
or, (x – a) cota
+ (y – b) = r [sina + cosa ×
cota]
or, (x – a) cosa
+ (y – b) sina = r
Alternate:
Equation of circle
(x–a)2 + (y – b)2 = r2
Tangent at (x,y) = (a +
rcosa, b + rsina) is
(x – a) (r cosa
+ a – a) + (y – b) (r sina + b – b) = r2
&⇒
(x – a) cosa + (y – b)sina
= r
Example
2.
From the
point A (0, 3) on the circle x2 + 4x + (y - 3)2 = 0 a
chord AB is drawn and extended to a point M, such that AM = 2AB. Find the
equation of locus of M.
Solution
Here AM = 2AB
&⇒
AB = 
&⇒
B is the mid point of AM. Let M be (h, k).
Hence B º
But AB is the chord of
the circle x2 + 4x + (y – 3)2 = 0 and point B lies on
the circle. .
Hence 
+
4 × 
+
or,
or, 
or,
h2 + 8h + k2 + 9 – 6k = 0
or, h2
+ k2 + 8h – 6k + 9 = 0
&⇒ Locus of M is x2 + y2
+ 8x – 6y + 9 = 0
Example
3.
Let A be the
centre of the circle. x2 + y2 - 2x - 4y - 20 = 0. Suppose
that the tangents at the points B (1, 7) and D (4, -2) on the circle meet at
the point C. Find the area of the quadrilateral ABCD.
Solution
Equation of the circle is x2
+ y2 – 2x – 4y – 20 = 0 where centre is A (1, 2) and
|
radius = AB
= AD = 
Equation of
the tangent at B (1, 7) is
x + 7y – (x +
1) – 2(y + 7) – 20 = 0
Or x + 7y – x
– 1 – 2y – 14 – 20 = 0
&⇒
5y = 35
&⇒ y = 7 … (1)
Equation of
the tangent at D (4, –2) is
4x – 2y – (x +
4) – 2(y – 2) – 20 = 0
or 4x – 2y – x
– 4 – 2y + 4 – 20 = 0
|
|
or 3x – 4y – 20 = 0 …(2)
Point of intersection of line (1) and
(2) is C given by
3x – 28 – 20 = 0 or, 3x = 48 or, x
= 16
C (16, 7), BC = 
=
DC = 15
Area of quadrilateral ABCD = Area D
ABC+ Area of D ADC
= 
=
75. Sq.units.
Example
4.
The cente of
a circle is (1, 1) and its radius is 5 units. If the centre is shifted along
the line y – x = 0 through a distance Ö2 units. Find the equation of the
circle in the new position. .
Solution
y = x has slope tanq
= 1, q = 
Hence 
So centre in the new position are
(2, 2), (0, 0). .
Equation of circle are x2
+ y2 = 25 and (x – 2)2 + (y – 2)2 = 25
Example
5.
A variable
circle passes through the point A (a, b) and touches the x - axis. Show that
the locus of the other end of the diameter through A is (x - a)2 =
4by.
Solution
Let the other end of
the diameter be(x1, y1). The centre and the radius of the
required circle are given by C
and
Radius = 
&⇒
|
&⇒
&⇒
&⇒
&⇒
(x1–a)2=4by1. .
Hence the
locus of (x1, y1) is
&⇒
(x - a)2 = 4by
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