# Circles - 2

(7) __Parametric form__-

general point of a circle if centre is (0,0) isparameter (radius)

__Illustration -2__- Find the equation a circle which touches the .y axis at (0, 4) and cuts an intercept of length .6 units on x axis .

__Solution__- The equation of circle toching x = 0 at (0,4) can be taken as (x - 0)^{2} + (y - 4^{2} ) + kx = 0

x^{2} + y^{2} + kx - 8y + 16 = 0

the circle cuts x -axis point (x_{1}, 0) .8 (x_{2}, 0 )given by, x^{2} + kx + 16 = 0

Xintercept difference of root of this quadratic equation 6 = | x_{2} - x_{1}|

36 = (x_{2} + x_{1})^{2} - 4x_{1} .x_{2}

36 = k^{2}- 4 (16)

k = 10

Hence the required circle is ,

x^{2} + y^{2}10 x - 8y + 16 = 0

__Some natations in a circle__-

1) s = x^{2} + y^{2} + 2gx + 2fy + c

2) s_{1}= x x_{1}^{2} + y_{1}^{2} + 2yx_{1} + 2fy_{1} + c

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

__Standrad form__-

1)s = x^{2} + y^{2} - a ^{2}

2)s =_{1}x_{1}^{2} + y_{1} ^{2} - a^{2}

3) T = xx_{1} + yy_{1} - a2

* If s_{1}> 0 point lies out side the circle

* If s_{1}< 0 point lies in side the circle

* If s_{1} 0 point lies upon the circle

__Why __:-

Let equation of circle be X^{2} + y^{2} + 2yx + 2fy + c = 0

having centre C( -y, -f) and radius

Let P (x_{1}, y_{1}) be any point then :-

P lies outside the circle if :-

PC > r

=> x_{1} ^{2} + y_{1}^{2} +2 yx _{1} + 2fy_{1} + c > 0

P lies on the circle if :-

PC = r

=> x_{1} ^{2} + y_{1}^{2} +2gx_{1} +2fy _{1} + c = 0

P lies inside the circle if :-

PC < r

=> x_{1}^{2} + y_{1}^{2} +2gx_{1} 2fy_{1} + c < 0

__ Dumb question __:-

How does PC > r leads to -

x_{1}^{2}+ y_{1}^{2} + 2gx_{1} + 2fy_{1} + c > 0

__Ans__-

and r =

Now PC > r

=> PC^{2} > r^{2}

=> (x_{1} +y)^{2}+(y_{1} +f)^{2} > y^{2} + f^{ 2}- c

=> x_{1}^{2} + y_{1}^{2} 2gx, + 2fy_{1} + c > 0

(1) A line L and a circle intersed in two point A and B .

=> d < r

=> Perpendi cular distance of line L from the centre of circle is less than the radius, and the length of te chors AB is :-

(2) A line L and +a circle touch each other at a point P.

=> d = r

=> Perpendicular distance of L from the centre of circle = radius.

(3) A line L and a circle may not intersect at all

=> .d > r

=> Perpendicular distance of line from the centre of circle is greater than the radius .

(4) A line y = mx + c touches circle x^{2} + y^{2} = a^{2}

If :- perpendicular distance of line from centre of the circle

= radius of the circle

__Illustration__- For what value of m, will line y = mx does not intersect the circle x^{2}+y^{2} + 20 X +20y + 20 = 0

__Solution__- IF the line y = mx does not intersect the circle ; the perpendicular distance of the line from the centre of the circle must be greater than its radius .

Centre of circle (-10, -10) ; radius

distance of line mx - y = 0 from (-10, -10)

=> |m(-10)-(-10)|

=>(2m + 1) (m + 2) < 0

=> -2 < m < - ^{1}⁄_{2}

__Intersection of line with circle-__

Let the line be y = mx + d and circle is x^{2}+ y^{2} +2gx + 2fy + c

thes x. Coordinate of their point of intersection are given by, (1 + m^{2})x2+ (2g + 2fm +2dm) x + d^{2}+ 2fd + = o

Why :-

When the two curves intersect, both the curves will be simultaneously satisfied.

So y = mx + d can be replaced in

x^{2}y^{2} + 2gx + 2fy + c =0

=> x^{2} + (mx + d )^{2} + 2gx + 2f (mx + d) + c =0

=> (1 + m)^{2} + (2g + 2fm + 2dm) x + d^{2} + 2fd + c = 0

__if.__ (i) B^{2} - 4AC = 0 then line touches the circle.

(ii) B^{2} - 4AC = > 0 then the line intersect circle at 2 different point.

(iii) B^{2} - 4AC = < 0 then no real intersecti takes place.

__Illustration 4__- Find the point on the circle x^{2} + y^{2}

= 4 whose distance from the line 4x + 3y = 12 is ^{4}⁄_{5} .

__Soluction- __Let A,B be the point on x^{2}/u> + y^{2} = 4 luing ar a distance ^{4}⁄_{5} from 4x + 3y = 12

=> AB will be parallel to 4x + 3y = 12

distance between the two line is

=> C = 16, 8

=> the equation of AB is :- 4x + 3y = 8 4x + 3y = 16

the point A,B can be formed by sliving for point of intersection of x^{2} y^{2} = 4 with AB.

AB(4x + 3y - 8 = 0)

=>

=> 25 x^{2} - 64x + 28 = 0

=> x = 2, ^{14}⁄_{25}

y = 0, ^{48}⁄_{25}

AB (4x + 3y - 16 = 0)

=> => 25 x^{2} - 128 x + 220 = 0

=> D < 0 => no real roots

Hence these are two pointr on circle at distance ^{4}⁄_{5} from liine

A(2,0) . & B (^{14}⁄_{25}, ^{48}⁄_{25})