Write the equation of the circle. Center at (4,2) and the circle passes through the point (-1,-1).
Please help!
recall the equation of a circle with centre (h,k) and radius r is
(x-h)^2 + (y-k)^2 = r^2
so
(x-4)^2 + (y-2)^2 = r^2
but (-1,-1) lies on it
sub x=-1 and y=-1 to find r^2
let me know what you get.
To find the equation of a circle, we need to know the coordinates of its center and the radius. In this case, we are given the center at (4, 2) and a point that lies on the circumference of the circle, (-1, -1).
The distance formula can be used to find the radius of the circle. The distance between two points (x1, y1) and (x2, y2) is given by:
d = √((x2 - x1)^2 + (y2 - y1)^2)
In this case, the distance from the center (4, 2) to the point (-1, -1) is the radius of the circle. So we have:
r = √((-1 - 4)^2 + (-1 - 2)^2)
= √((-5)^2 + (-3)^2)
= √(25 + 9)
= √34
Now that we have the radius (√34) and the center (4, 2), we can write the equation of the circle in standard form as:
(x - h)^2 + (y - k)^2 = r^2
where (h, k) represents the coordinates of the center. Substituting the given values, we have:
(x - 4)^2 + (y - 2)^2 = (√34)^2
(x - 4)^2 + (y - 2)^2 = 34
Therefore, the equation of the circle is (x - 4)^2 + (y - 2)^2 = 34.