Find the equation of the circle with centre on the line 2x + y = 0 and touching the lines 4x - 3y + 10 = 0 and 4x - 3y - 30 = 0.

Since the two lines are tangents to the given circle and they are parallel, we have

Note: We know that if ax + by + c = 0 and ax + by + c¢ = 0are two parallel tangents to a circle, then equation of the line parallel to given lines and passing through the centre is given by ax + by + .

Hence the centre of the circle lies on 4x - 3y - 10 = 0

Also the centre lies on 2x + y = 0

Hence the coordinates of the centre are (1, -2).

&⇒
The equation of the circle is (x - 1)^{2} + (y + 2)^{2} = 16.

Find the
locus of the mid-points of the chords of the circle

x^{2} + y^{2} - 2x - 6y - 10 = 0 which pass through the origin.

Let (h, k) be the coordinates of the mid point.

Equation of the chord whose mid point is (h, k) is

xh + yk - (x + h) - 3(y + k) - 10 = h^{2}
+ k^{2} - 2h - 6k - 10 (using T = S_{1})

i.e., h^{2} + k^{2} -
h(x + 2) - k(6 + y) + x + h + 3y + 3k = 0.

This chord passes through the origin (0, 0). .

Hence h^{2} + k^{2} -
h - 3k = 0

Hence the required locus is x^{2}
+ y^{2} - x - 3y = 0.

Alternative:

If m_{1} is the slope of the
chord, then

If m_{2} is
the slope of the perpendicular to the chord then where
(1, 3) is

the centre of the given
circle. But m_{1}m_{2} = -1 .

&⇒
h^{2} + k^{2} - h - 3k = 0

Hence the required locus is x^{2}
+ y^{2} – x – 3y = 0.

Find the
equations of the tangents from the point A(3, 2) to the circle

x^{2} + y^{2} = 4 and hence find the angle between the pair of
tangents.

The
equation of the pair of tangents from (3, 2) is given by T i.e.,
9x Here a = 0, b = 5 h = -6. . Hence angle q between the tangents is given by or tanq = or
tanq = Hence the angle between the pair of tangents is tan |

Find the locus
of middle points of chords of the circle (x–2)^{2} + (y –3)^{2}
= a^{2}, which subtends a right angle at the point (b, 0).

Let M (h, k) be the middle points of the chord AB. Also ∠APB = 90° and AM = PM. We have AM = = PM = Hence locus of (h, k) is |

x^{2} + y^{2} – x (b
– 2) – 3y +

which is a circle.

If a straight line through C(-Ö8, Ö8), making an angle of 135° with the x-axis, cuts the circle x = 5 cos q, y = 5 sin q, in points A and B, find the length of the segment AB.

The given
circle is x^{2} + y^{2} = 25 . . .(1).

Equation of line through C is

&⇒

Any point on this line is

If the point
lies on the circle x^{2} + y^{2} = 25, the

&⇒

&⇒

&⇒ r^{2}
+ 8r + 16 – 25 = 0

&⇒ r^{2} – 9r + r – 9 = 0

&⇒
r = 9, –1

AB = 9 – (–1) = 10.