Find an equation of the circle that has center (-3,6) and passes through (4,-1)
recall that the circle with center at (h,k) and radius r has equation
(x-h)^2 + (y-k)^2 - r^2
The radius here is the distance from the center to (4,-1).
Now just plug in your numbers.
To find the equation of a circle, we need to know the center coordinates and either the radius or a point that lies on the circle. In this case, we have the center point (-3,6) and a point on the circle (4,-1).
The general equation of a circle with center (h, k) and radius r is:
(x - h)^2 + (y - k)^2 = r^2
Substituting the given values, we have:
(x - (-3))^2 + (y - 6)^2 = r^2
Simplifying:
(x + 3)^2 + (y - 6)^2 = r^2
Now, we need to find the radius (r). We can use the distance formula to find the distance between the center (-3,6) and the point on the circle (4,-1).
The distance formula is given by:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Substituting the values:
d = sqrt((4 - (-3))^2 + (-1 - 6)^2)
d = sqrt(7^2 + (-7)^2)
d = sqrt(49 + 49)
d = sqrt(98)
Now, we have the radius (r), which is equal to d:
r = sqrt(98)
Substituting this back into the equation of the circle:
(x + 3)^2 + (y - 6)^2 = (sqrt(98))^2
(x + 3)^2 + (y - 6)^2 = 98
Hence, the equation of the circle is (x + 3)^2 + (y - 6)^2 = 98.