(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 - 42 ) + kx = 0
x2 + y2 + kx - 8y + 16 = 0
the circle cuts x -axis point (x1, 0) .8 (x2, 0 )given by, x2 + kx + 16 = 0
Xintercept difference of root of this quadratic equation 6 = | x2 - x1|
36 = (x2 + x1)2 - 4x1 .x2
36 = k2- 4 (16)
k = 10
Hence the required circle is ,
x2 + y210 x - 8y + 16 = 0
Some natations in a circle-
1) s = x2 + y2 + 2gx + 2fy + c
2) s1= x x12 + y12 + 2yx1 + 2fy1 + c
3) T = xx1 + yy1 + g(x + x1) + f (y + y1) + c
1)s = x2 + y2 - a 2
2)s =1x12 + y1 2 - a2
3) T = xx1 + yy1 - a2
* If s1> 0 point lies out side the circle
* If s1< 0 point lies in side the circle
* If s1 0 point lies upon the circle
Let equation of circle be X2 + y2 + 2yx + 2fy + c = 0
having centre C( -y, -f) and radius
Let P (x1, y1) be any point then :-
P lies outside the circle if :-
PC > r
=> x1 2 + y12 +2 yx 1 + 2fy1 + c > 0
P lies on the circle if :-
PC = r
=> x1 2 + y12 +2gx1 +2fy 1 + c = 0
P lies inside the circle if :-
PC < r
=> x12 + y12 +2gx1 2fy1 + c < 0
Dumb question :-
How does PC > r leads to -
x12+ y12 + 2gx1 + 2fy1 + c > 0
and r =
Now PC > r
=> PC2 > r2
=> (x1 +y)2+(y1 +f)2 > y2 + f 2- c
=> x12 + y12 2gx, + 2fy1 + 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 x2 + y2 = a2
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 x2+y2 + 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)
=>(2m + 1) (m + 2) < 0
=> -2 < m < - 1⁄2
Intersection of line with circle-
Let the line be y = mx + d and circle is x2+ y2 +2gx + 2fy + c
thes x. Coordinate of their point of intersection are given by, (1 + m2)x2+ (2g + 2fm +2dm) x + d2+ 2fd + = o
When the two curves intersect, both the curves will be simultaneously satisfied.
So y = mx + d can be replaced in
x2y2 + 2gx + 2fy + c =0
=> x2 + (mx + d )2 + 2gx + 2f (mx + d) + c =0
=> (1 + m)2 + (2g + 2fm + 2dm) x + d2 + 2fd + c = 0
if. (i) B2 - 4AC = 0 then line touches the circle.
(ii) B2 - 4AC = > 0 then the line intersect circle at 2 different point.
(iii) B2 - 4AC = < 0 then no real intersecti takes place.
Illustration 4- Find the point on the circle x2 + y2
= 4 whose distance from the line 4x + 3y = 12 is 4⁄5 .
Soluction- Let A,B be the point on x2/u> + y2 = 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 x2 y2 = 4 with AB.
AB(4x + 3y - 8 = 0)
=> 25 x2 - 64x + 28 = 0
=> x = 2, 14⁄25
y = 0, 48⁄25
AB (4x + 3y - 16 = 0)
=> => 25 x2 - 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)
(7) Parametric form-