The point (-4.3) lies on a circle whose center is at (2,1). Find an equation of this circle.
start with the equation
(x-2)^2 + (y-1)^2 = r^2
sub in the point (-4,3) to find r^2
Is the answer
(x-2)^2+(y-1)^2=40
correct
To find the equation of a circle, you need to know the coordinates of its center and either its radius or any point that lies on the circle. In this case, you're given the coordinates of the center, (2,1), and a point on the circle, (-4,3).
To find the radius of the circle, you can use the distance formula between the center and the point on the circle. The distance formula is given by:
distance = √((x2 - x1)^2 + (y2 - y1)^2)
Using the given points, the distance between the center (2,1) and the point on the circle (-4,3) is:
distance = √((-4 - 2)^2 + (3 - 1)^2) = √((-6)^2 + (2)^2) = √(36 + 4) = √40
So, the radius of the circle is √40.
Now, the equation of a circle with center (h, k) and radius r is given by:
(x - h)^2 + (y - k)^2 = r^2
Substituting the values of the center (h, k) = (2,1) and the radius r = √40, we get:
(x - 2)^2 + (y - 1)^2 = 40
Therefore, the equation of the circle is (x - 2)^2 + (y - 1)^2 = 40.