Find the
equation of the tangent at A to the circle ( x - a)^{2} + (y - b)^{2}
= r^{2}, where the radius through A makes an angle a
with the x - axis.

The parametric equations of the circle

(x–a)^{2} + (y
– b)^{2} = r^{2} 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} = r^{2}

Tangent at (x,y) = (a + rcosa, b + rsina) is

(x – a) (r cosa
+ a – a) + (y – b) (r sina + b – b) = r^{2}

&⇒
(x – a) cosa + (y – b)sina
= r

From the
point A (0, 3) on the circle x^{2} + 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.

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 x^{2} + 4x + (y – 3)^{2} = 0 and point B lies on
the circle. .

Hence + 4 × + or,

or, or,
h^{2} + 8h + k^{2} + 9 – 6k = 0

or, h^{2}
+ k^{2} + 8h – 6k + 9 = 0

&⇒ Locus of M is x^{2} + y^{2}
+ 8x – 6y + 9 = 0

Let A be the
centre of the circle. x^{2} + y^{2} - 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.

Equation of the circle is x^{2}
+ y^{2} – 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 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.

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. .

y = x has slope tanq = 1, q =

Hence

So centre in the new position are (2, 2), (0, 0). .

Equation of circle are x^{2}
+ y^{2} = 25 and (x – 2)^{2 }+ (y – 2)^{2} = 25

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.

Let the other end of
the diameter be(x_{1}, y_{1}). The centre and the radius of the
required circle are given by Cand
Radius =

&⇒

Hence the
locus of (x |