IIT JEE

- If S º x
^{2}+ y^{2}+ 2gx + 2fy + c = 0 and S¢ º x^{2}+ y^{2}+ 2g¢x + 2f¢y + c¢ = 0 are two intersecting circles, then S + lS¢ = 0, l > -1, is the equation of the family of circles passing through the points of intersection of S = 0 and S¢ = 0. - If S º x
^{2}+ y^{2}+ 2gx + 2fy + c = 0 is a circle which is intersected by the straight line

m º ax + by + c = 0 at two real and distinct points, then S + lm = 0 is the equation of the family of circles passing through the points of intersection of S = 0 and m=0.

Ifm = 0 touches S = 0 at P, then S + lm = 0 is the equation of the family of circles, each touching m = 0 at P. - The equation of a family of circles
passing through two given points (x
_{1}, y_{1}) and

(x_{2}, y_{2}) can be written in the form - (x - x
_{1})(x - x_{2}) + (y - y_{1})(y - y_{2}) + where l is a parameter. - The equation of the family of circles
which touch the line y - y
_{1}= m(x - x_{1}) at

(x_{1}, y_{1}) for any value of m is (x - x_{1})^{2 }+ (y - y_{1})^{2}+ l[(y - y_{1}) -m(x - x_{1})] = 0. If m is infinite, the equation is (x - x_{1})^{2}+ (y - y_{1})^{2}+ l(x - x_{1}) = 0. - The two circles are said to intersect orthogonally if the angle of intersection of the circles i.e., the angle between their tangents at the point of intersection is 90° .

The condition for the
two circles S = 0 and S_{1}= 0 to cut each other orthogonally is 2gg_{1}
+ 2ff_{1} = c + c_{1 }. .

Find
the equation of the circle described on the common chord of the circles x^{2}
+ y^{2} - 4x - 5 = 0 and x^{2} + y^{2} + 8y + 7 = 0 as
diameter.

Equation of the common chord is S_{1}
- S_{2} = 0

&⇒ x + 2y + 3 = 0

Equation of the circle
through the two circles is S_{1} + lS_{2} =
0

&⇒
x^{2} + y^{2} - x + + = 0

Its centre lies on x + 2y + 3 = 0

&⇒
+ 3 = 0

&⇒
2 - 8l + 3 + 3l = 0

&⇒
l
= 1. Hence the required circle is x^{2} + y^{2} - 2x + 4y + 1
= 0.

Show
that the circle passing through the origin and cutting the circles x^{2}
+ y^{2} - 2a_{1}x - 2b_{1}y + c_{1} = 0 and x^{2}
+ y^{2} - 2a_{2}x - 2b_{2}y + c_{2} = 0
orthogonally is = 0 .

Let the equation of the circle passing through the origin be

x^{2} + y^{2} + 2gx +
2fy = 0, . . . . (1).

It cuts the given two circles
orthogonally

&⇒ -2ga_{1} - 2fb_{1} =
c_{1}

&⇒
c_{1} + 2ga_{1} + 2fb_{1} = 0, .
. . . . (2).

and -2ga_{2} - 2fb_{2}
= c_{2}

&⇒
c_{2} + 2ga_{2} + 2fb_{2} = 0. .
. . . . . (3).

Eliminating 2f and 2g from (1), (2) and (3), we get

= 0 .

Tangents PQ and PR
are drawn to the circle x^{2} + y^{2} = a^{2} from the
point P(x_{1}, y_{1}). Prove that equation of the circum circle
of DPQR is

x^{2 }+ y^{2} –xx_{1} – yy_{1}= 0.

QR is the chord of contact of the tangents to the circle

x^{2} + y^{2} – a^{2}
= 0 ….(1).

equation of QR is xx_{1} + yy_{1}
– a^{2} ….(2) .

The circumcircle of DPQR
is a circle passing through the intersection of the circle (1) and the line (2)
and the point P(x_{1}, y_{1}). .

Circle through the intersection of (1) and (2) is

x^{2} + y^{2} – a^{2}
+ l (xx_{1} + yy_{1} –a^{2}) = 0 …(3)

it will pass through (x_{1},
y_{1}) if

x_{1}^{2} + y_{1}^{2}
– a^{2} + l (x_{1}^{2} + y_{1}^{2}
–a^{2}) = 0

l
= –1 (since x_{1}^{2} + y_{1}^{2} >
a^{2})

Hence equation of circle is (x^{2}
+ y^{2} – a^{2}) – (xx_{1} + yy_{1} –a^{2})
= 0

Or x^{2} + y^{2} –
xx_{1} – yy_{1} = 0 . .