Find the coordinates of
the points at which the circles x^{2} + y^{2} - 4x - 2y = 4 and

x^{2} + y^{2} - 12x - 8y = 12 touch each other. Find the
coordinates of the point of contact and equation of the common tangent at the
point of contact.

The given circles are

S_{1}
º
x^{2} + y^{2} – 4x – 2y – 4 = 0 and S_{2} º
x^{2} + y^{2} – 12x – 8y – 12 = 0

Equation of
the common tangent is S_{1} – S_{2} = 0 i.e. 8x + 6y + 8 = 0.

&⇒ 4x + 3y + 4 = 0 . . .
(1).

Let the
point of contact be (x_{1}, y_{1})

Equation of
the tangent at (x_{1}, y_{1}) is ,

xx_{1}
+ yy_{1} – 2(x + x_{1}) – (y + y_{1}) – 4 = 0

or (x_{1}
– 2)x + (y_{1} – 1)y – 2x_{1} – y_{1} – 4 = 0 .
. . (2).

Comparing (1) and (2), we get

&⇒ x_{1} = 4k + 2, y_{1}
= 3k + 1, 2x_{1} + y_{1} + 4 = – 4k

&⇒ 2(4k + 2) + 3k + 1 + 4 = – 4k or
8k + 4 + 3k + 5 = – 4k

&⇒ 15k = –9

&⇒
k =

&⇒
x_{1} = 4 ´ ,
y_{1} = 3 ´

Point of contact =

A variable triangle has
two of its sides along the axes. Its 3rd side touches the circle x^{2}
+ y^{2} - 2ax - 2ay + a^{2} = 0. Prove that the locus of the
circumcentre of the triangle is a^{2} - 2a(x + y) + 2xy = 0.

The given
circle has its centre (a, a) and radius a, so that the circle touches both
the axes. Let the side AB be

&⇒
A º (p, o), B º (o, q)

Let R be the midpoint of AB, so that

R º
=
(h, k)

&⇒ p = 2h, q = 2k

Since the line touches the circle,

So, (ap + aq –
pq)
Locus of (h, k) is a |

Consider a curve ax^{2}
+ 2hxy + by^{2} = 1, and a point P not lying on the curve. A line drawn
from the point P intersects the curve at points Q and R. If the product PQ.PR
is independent of the slope of the line, show that the curve is a circle.

Let P be the point (a, b). The equation of the line through P, with slope tanq, is = r

Any point on this line is (a + r cosq, b + r sinq). If this point lies on the curve, then .

a(a
+ r cosq)^{2} + 2h(a + r cosq)
(b
+ r sinq) + b(b + r sinq)^{2}
= 1

This is a quadratic equation in r
whose values give the coordinates of Q and R, where PQ = r_{1} and PR =
r_{2}.

&⇒
r_{1}r_{2} =

=

which is independent of
q
if +
h^{2} = 0

&⇒ a = b, h = 0.

Hence the given curve is a circle.

Find the equation of
the circle passing through (-4, 3) and touching the lines

x + y = 2 and x - y = 2.

The lines are mutually perpendicular and passes through (2, 0). . Due to symmetry, obviously the centre of the required circle lies on x-axis. Let it be (-a, 0). So equation of circle is x It passes through (-4, 3) \ 16 + 9 – 8a + c =
0 |

radius =

But radius = length of perpendicular from (-a, 0) on x + y = 2. .

This gives a = 10 ± 3

&⇒
c = 55 ± 24

circle is x^{2} + y^{2}
+ 2(10 ± 3)x
+ 55 ± 24=
0

Find the
circumcentre of the triangle formed by x + y =0, x – y = 0 and lx + my =1. If l
and m vary such that l^{2} + m^{2} = 1, show that the locus of
its circumcentre is the curve (x^{2} – y^{2})^{2} = x^{2}
+ y^{2} .

Solving the straight lines x – y = 0, x + y = 0, lx + my = 1

we get the vertices as A(0, 0), Band

if S(h, k) is the circumcentre then
SA^{2} = SB^{2} = SC^{2} and hence

h^{2}_{ }+ k^{2}
= =

or h + k = ….(1)

or h – k = ….(2)

if l, m vary such that l^{2}
+ m^{2} = 1, then the locus of the circumcentre is got from

h^{2}_{ }+ k^{2}
= =
(h^{2} – k^{2})^{2}

Hence locus of circumcentre is (x^{2}
– y^{2})^{2} = x^{2} + y^{2} . .