A circle of diameter 13 m with the centre O coinciding with the origin of co-ordinate axes, has the diameter AB on the x-axis (x co-ordinates of B > 0). If the length of the chord AC be 5m, find the equation of pair of lines BC, C having two possible positions. .

Given AB = 13m, AC = 5m. Let OL ^AC, then L is the mid-point of AC. .

Now AL =and AO =

From D ALO, cosq =

\ slope of BC =tan(90° + q) = – cotq =

Again slope of BC¢

= tan(– 90° – q) = cotq =

\ Equation of BC and BC¢ are

or y

and or

\their
joint equation is y^{2} -

or

or 576y^{2} –100x^{2}
+1300x –4225 = 0

100x^{2} – 576y^{2}
–1300x + 4225 = 0 .

If S_{1}
º
x^{2} + y^{2} +2g_{1}x + 2f_{1}y +c_{1}
= 0 and S_{2} º x^{2} + y^{2} +2g_{2}x
+ 2f_{2}y +c_{2} = 0 are two circles with radii r_{1 }and
r_{2} respectively, show that the points at which the circles subtend
equal angles lie on the circle .

If the angle subtended
by the circles S_{1} = 0 and S_{2} = 0 at point A are q
then

∠SAR
= ∠PAQ = q

\
∠C_{1}AR
= ∠PAC_{2} =

Let A(h, k). Then AR = , AP = .

&⇒

&⇒

\ locus of (h, k) is i.e. .

A chord of
the circle x^{2} + y^{2} - ax - by = 0, drawn from the point
(a, b), is divided by the x-axis in the ratio 2 : 1. Prove that a^{2 }>
3b^{2}. .

The given point A(a, b) lies on the given
circle. Any point on the circle is P.
The x-axis cuts the chord AP in the ratio 2:1. Hence =
0

&⇒ sinq = -

&⇒ |sinq| = <
1

&⇒ 4b^{2} <
a^{2} + b^{2}

&⇒
a^{2} > 3b^{2 }.

Find the equation to
the straight lines joining the origin to the points in which the straight line
y = mx +c cuts the circle x^{2} + y^{2} =2ax + 2by. .

Hence find the condition that these points may subtend a right angle at the origin. Find also the condition that the straight line may touch the circle.

The straight line is given as y =mx + c or ….(1)

The equation of the circle is given as

x^{2} + y^{2} = 2ax +
2by or x^{2} + y^{2} –2ax – 2by =0 ….(2).

Making the second equation homogeneous with the help of (1), we get the combined equation of the straight lines joining the points of intersection of (1) with (2) to the origin, hence the required equation is

x^{2} + y^{2} –2 ax= 0

or cx^{2} + cy^{2}
–2ax(y – mx) – 2by(y –mx) = 0

or x^{2} ( c+2am) + y^{2}
(c –2b) –2xy(a –bm) = 0 . . . (3).

If the lines represented by this
equation are at right angles, the sum of the coefficients of x^{2} and
y^{2} should be zero. Hence the required condition is

c + 2am + c –2b = 0 or c = b – am.

If the line (1) touches the circle, it isonly possible when the line cuts the circle in two coincident points. Therefore the lines joining the origin to the two points will also coincide. Hence the equation (3) must represent one line only. It is only possible when it is a perfect square, so its discriminant must be zero. .

So [ 2 (a – bm)]^{2} –4(c
+2am) (c – 2b) = 0

Or a^{2} + b^{2}m^{2}
– 2abm –c^{2} +2bc –2amc + 4abm = 0

Or c^{2} – 2c (b – am) –a^{2}
- b^{2}m^{2}_{ }– 2abm = 0

Or c^{2} – 2c (b –am) – (a +
bm)^{2} =0

\ c =

= b – am ±
Ö
[ ( b^{2} + a^{2} ) ( 1+m^{2 })] . .

DABC has vertices A(a cosa, a sina), B( a cosb, a sinb) and C(a cosg, a sing). Prove that its orthocentre is given by .

H{a(cosa + cosb + cosg), a(sina + sinb + sing)}. .

The DABC is
inscribed in circle x^{2} + y^{2} = a^{2}

so circumcentre is origin O

centroid is

Centroid divide line going circumcentre and orthocentre (h, k) in ratio 1 : 2. .

\

&⇒
h = a(cosa + cosb + cosg)

Similarly k = a(sina + sinb + sing).