Posted on - 23-01-2017

IIT JEE

If S = 0 be a curve then S_{1 }=0
indicate the equation which is obtained by substituting x= x_{1}
and y = y_{1} in the equation of the given curve, and T = 0 is the
equation which is obtained by substituting x^{2} = xx_{1} ,
y^{2} = yy_{1}, 2xy = xy_{1} + yx_{1},

2x = x+x_{1}, 2y=y+y_{1} in the equation S = 0. .

If S º x^{2}
+y^{2} +2gx +2fy + c = 0 then S_{1}º
x_{1}^{2} +y_{1}^{2} +2gx_{1} + 2fy_{1}
+c , and

T º xx_{1}+yy_{1}
+ g( x+x_{1}) + f(y +y_{1}) +c

- Equation of the tangent to x
^{2}+ y^{2}+ 2gx + 2fy + c = 0 at A(x_{1}, y_{1}) is

xx_{1}+ yy_{1}+ g(x + x_{1}) + f(y + y_{1}) + c = 0. - The condition that the straight line
y = mx + c is a tangent to the circle

x^{2}+ y^{2}=a^{2}is c^{2}= a^{2}(1 + m^{2}) and the point of contact is (-a^{2}m/c, a^{2}/c) i.e.

y = mx ± is always a tangent to the circle x^{2}+ y^{2}= a^{2}whatever be the

value of m. - The joint equation of a
pair of tangents drawn from the point A(x
_{1}, y_{1}) to the circle

x^{2}+ y^{2}+ 2gx + 2fy + c = 0 is T^{2}= SS_{1}. - The equation of the
normal to the circle x
^{2}+ y^{2}+ 2gx + 2fy + c = 0 at any point

(x_{1}, y_{1}) lying on the circle is . - In particular, equations of the
tangent and the normal to the circle x
^{2}+ y^{2}= a^{2}at

(x_{1}, y_{1}) are xx_{1}+ yy_{1}= a^{2}; and respectively. - The equation of the chord of the
circle S º 0, whose mid point (x
_{1}, y_{1}) is T = S_{1}. - The length of the
tangent drawn from a point (x
_{1}, y_{1}) outside the circle S º 0, to the circle is .

From a
point P(x |

Find the equation of the circle whose centre is (3, 4) and which touches the line 5x + 12y = 1.

Let r be the radius of the circle. Then.

r = distance of the centre i.e. point (3, 4) from the line 5x + 12y = 1.

=

Hence the equation of
the required circle is (x - 3)^{2} + (y - 4)^{2} =

&⇒
x^{2} + y^{2} - 6x - 8y + =
0.

Find
the co-ordinates of the point from which tangents are drawn to the circle x^{2}
+ y^{2} - 6x - 4y + 3 = 0 such that the mid point of its chord of
contact is (1, 1).

Let the required point
be P(x_{1}, y_{1}). The equation of the chord of contact of P
with respect to the given circle is.

xx_{1} + yy_{1}
- 3(x + x_{1}) - 2(y + y_{1}) + 3 = 0 (1)

The equation of the chord with mid-point (1, 1) is

x + y - 3(x + 1) - 2(y
+ 1) + 3 = 1 + 1 - 6 - 4 + 3

&⇒ 2x + y = 3

Equating the ratios of the
coefficients of x, y and the constant terms and solving for x, y we get x_{1}
= -1, y_{1} = 0.