Find the locus of the
point of intersection of tangents to the circle x = acos q,

y = asinq at the point whose parametric angles differ by (i) p/3
(ii) p/2.

Let A and B be two points on circle whose parametric angle differs by q

then PA = tan . . . . . (1) (i) when q = p/3 then eq.
(1) becomes h locus of point
of intersection P is x (ii) when q = p/2 eq.(1) becomes . h |

locus of point of
intersection P is x^{2} +y^{2} = 2a^{2}

A point moves so that the sum of the squares of its distances from n fixed point is given. Prove that its locus is a circle.

Let (h, k) be the
moving point and (a_{i}, b_{i}); i = 1, 2, 3, …, n be the fixed
points. Then given (constant)

&⇒

&⇒
n ( h^{2} + k^{2}) – 2h Sa_{i} – 2kSb_{i}
+ S(a_{i}^{2} + b_{i}^{2})
- l = 0

\
locus of (h, k) is x^{2} + y^{2} - 2a
x - 2b y + g = 0 which is a circle

where a = , b = , g =

Lines 5x + 12y - 10 =
0 and 5x - 12y - 40 = 0 touch circle C_{1} of diameter 6. If centre of
C_{1 }lies in the first quadrant, find the equation of circle C_{2},
which is concentric with circle C_{1} and cuts an intercept of length 8
on these lines.

Since the lines 5x +12y – 10 = 0 and 5x–12y– 40 =0 touch the circle, its centre lies on one of the angle bisectors of the given lines. i.e 5x + 12y – 10 = ±(5x – 12y – 40)
But the
centre lies in first quadrant Since C |

&⇒ 25 + 12h – 10
= ± 39

&⇒ 12h = ± 39 – 15

&⇒ 12h = 24

&⇒
h = 2 also h = (not
possible)

centre of C_{1}
= (5, 2)

Let r be
radius of C_{2}

&⇒ r^{2} = OP^{2} + AP^{2}
= 3^{2} + 4^{2} = 5^{2 }or r = 5

Equation of
C_{2} is (x – 5)^{2} + (y – 2)^{2} = 5^{2}

&⇒
x^{2} + y^{2} – 10x – 4y + 4 = 0

A and B are two points on the x-axis and the y-axis respectively. Two circles are drawn passing through the origin and having centres at A and B respectively. Show that AB bisects the common chord of these circles.

Let A(a, 0) and B(0, b). Then the two circles are .

x^{2} + y^{2}
– 2ax = 0 …(1)

and x^{2} + y^{2}
– 2by = 0

Common chord is ax – by = 0 ….(2) .

Equation of circle
through (1) and (2) is x^{2} + y^{2} – 2ax + l(ax
– by) = 0

i.e. x^{2} + y^{2}
+ ax(l - 2) - l by = 0 .

This will be circle
with common chord (2) as diameter if centre lies
on (2) i.e.

&⇒
l
=

Mid-point of common chord is

i.e.

This lie on line AB ( equation ) if

which is true

so AB passes through mid-point of common chord.

Prove that the locus of
the centre of a circle which touches the circle

x^{2 }+ y^{2} – 6x – 6y + 14 = 0 externally and touches the
y-axis is given by

y^{2 }– 10x –6y + 14 = 0. .

Centre of given circle is (3, 3) and radius = 2. .

Let r = radius and (h, k) be the centre of other circle. .

r = length of perpendicular from (h, k) on x = 0 (y-axis)

\ r = h

Equation is x^{2} + y^{2}
– 2hx – 2ky – h = 0

Again r + 2 = distance between the centres

&⇒
h + 2 =

&⇒
(h + 2)^{2} = (h – 3)^{2} + (k - 3)^{2}

&⇒
k^{2} – 6k – 10h + 14 = 0

\
locus of centre (h, k) is y^{2} - 6k – 10x + 14 = 0.