I need help on 1 ac Text book section 21 to 23 Problerns 1
I need help on 1 a-c
Solution
coordinates of square are
( -1,-1) , (-4,5) , (2,8) , (5,2)
diameter of the circle is distance between two corners
distance between (-4,5 ) and (2,8)
d = sqrt (x2 - x1)^2 + (y2 - y1)^2
d = 6.71
hence radius of the circle is 6.71 /2 = 3.35
standard equation of circle is given by
(x-h)^2 + (y-k)^2 = r^2
where h,k is the centre of the circle
we know that circle passes therough mid points of the corners of the square
so finding 2 points through which teh circle passes
mid point of (-4.5) and (-1,-1) = (-5/2 , 2)
mid point of (5,2) and (2,8) = (7/2 , 5 )
centre of the circle would be midpoint of (-5/2 , 2) and (7/2 , 5 )
centre = (1/2 , 7/2)
h = 1/2 , k = 7/2
hence, equation of circle is
(x- 1/2)^2 + ( y - 7/2)^2 = 3.35^2
b) equation of line parallel to (-1,-1) and (5,2)
slope of this line is 1/2
slope of parallel line would be 1/2
hence , equation of line with slope 1/2 and point ( 1/2 , 7/2 )
using y = mx + b and plugging the values of m and x, y
hence equation is
y = 1/2 x +13/4
