A rectangle
ABCD is inscribed in a circle with a diameter lying along the line

3y = x + 10. If A and B are the points (-6, 7) and (4, 7) respectively, find
the area of the rectangle.

Let O (h, k) be the centre of circle. (h, k) also lies on the line 3y = x + 10 |

Now AB =,

OB = = and BD = .

In DADB, BD^{2}
= AD^{2} + AB^{2 }i.e. 164 = AD^{2} + 100.

&⇒
AD^{2} = 64

&⇒ AD = 8

Area of Rectangle ABCD = 10 ´ 8 = 80 sq. units.

AB is the diameter of a circle, CD is a chord parallel to AB and 2 CD = AB. The tangent at B meets the line AC produced at E. Prove that AE = 2AB.

Let O, the centre of circle, be taken as the origin. A (-a, 0) and B (a, 0) a is radius of the . circle. In DOCF,. OC Equation of the line AE where A (-a, 0), is y – 0 = |

y = Ö3 (x + a) … (1)

Equation of line BE [which is tangent at (a, 0) is

x = a … (2)

Point of intersection of (1) and (2) is E i.e. E (a, 2Ö3a).

AE = = 4a = 2(2a) = 2AB

Hence AE = 2AB

Find the
locus of centres of the circle which touches the two circle x^{2} + y^{2}
= a^{2} and x^{2} + y^{2} = 4ax externally. .

Let (h, k) be the centre of the circle and r be its radius then = r + a

and

Eliminating r,

\ locus of (h, k) is

Squaring (x – 2a)^{2} + k^{2}
= a^{2} + x^{2} + y^{2} + 2a

&⇒
- 4ax + 3a^{2} = 2a

Squaring we
get 12 x^{2} – 4y^{2} – 24ax + 9a^{2} = 0.

If i
= 1, 2, 3, 4 m_{i} > 0 are 4 distinct points on a circle, then show
that

m_{1} .m_{2} .m_{3} .m_{4} = 1.

where i = 1, 2, 3, 4 are four points lying on a circle.

Let
the equation of the circle be x^{2} + y^{2} + 2gx + 2fy + c = 0

is lying on the circle

&⇒ m_{i}^{2} + +
2gm_{i} + =
0

&⇒ m_{i}^{4} + 2gm_{i}^{3}
+ 2fm_{i} + cm_{i}^{2} + 1 = 0

If m_{1}, m_{2}, m_{3},
m_{4} are its roots then

&⇒ m_{1} ×
m_{2} × m_{3} ×
m_{4} = 1

If al^{2}
– bm^{2} + 2dl + 1 = 0 a, b, d are fixed real numbers such that a + b =
d^{2}, then prove that the line lx + my + 1 = 0 touches a fixed circle.
Find its equation.

Let the equation of the circle be x^{2}
+ y^{2} + 2gx + 2fy + c = 0

If lx + my + 1 = 0 is a tangent to the circle

Then

&⇒
(g^{2} + f^{2} – c) (l^{2 }+ m^{2}) = (lg + mf
– 1)^{2}

&⇒
(f^{2} – c) l^{2} + (g^{2} – c)m^{2} + 2l (g –
mgf) + 2mf – 1 = 0

Comparing with the given condition al^{2}
– bm^{2} + 2dl + 1 = 0,

– (f^{2} – c) = a, – (g^{2}
– c) = – b, -g (1 – mf) = d, f = 0

&⇒
c = a, g = – d, g^{2} – c = b, d^{2} – a = b

&⇒
a + b = d^{2} which is the required condition.

Hence the
fixed circle is x^{2}+ y^{2} – 2dx + a = 0.