The points B and C with coordinates (5,5) and (-3,1) respectively are the ends of the diameter of a circle centre A.
a)On the grid below draw a circle
b) determine the equation of the circle expressing it in the form x^2+y^2+ax+by+c=0 where a,b and c are integral values
a) To draw a circle with B and C as the ends of the diameter, we can find the midpoint of BC as the center of the circle.
The midpoint of BC is calculated as:
Midpoint:
x-coordinate: (x1 + x2) / 2 = (5 + (-3)) / 2 = 2 / 2 = 1
y-coordinate: (y1 + y2) / 2 = (5 + 1) / 2 = 6 / 2 = 3
So, the center of the circle is at point A coordinates (1, 3).
We can now draw the circle with center A and a radius of the distance from A to either B or C.
b) To find the equation of the circle, we can use the general equation of a circle:
(x - h)^2 + (y - k)^2 = r^2
where (h, k) are the coordinates of the center, and r is the radius of the circle.
From part a), we found the center of the circle at (1, 3). Now, we need to calculate the radius.
Radius:
Distance formula: √((x2 - x1)^2 + (y2 - y1)^2)
r = √((5 - 1)^2 + (5 - 3)^2) = √(16 + 4) = √20
Thus, the equation of the circle is:
(x - 1)^2 + (y - 3)^2 = 20
Expanding this equation, we get:
(x^2 - 2x + 1) + (y^2 - 6y + 9) = 20
x^2 + y^2 - 2x - 6y - 10 = 0
Therefore, the equation of the circle is:
x^2 + y^2 - 2x - 6y - 10 = 0, where a = -2, b = -6, and c = -10.