Knowing the given conditions, write the equation of the circle. Center at (4,2) and the circle passes through (-1,-1).
Please help!
I will assume you know that the equation of a circle with centre (h,k) and radius r is
(x-h)^2 + (y-k)^2 = r^2
then (x-4)^2 + (y-2)^2 = r^2
but (-1,-1) lies on it, so
(-1-4)^2 + (-1-2)^2 = r^2
r^2 = 34
(x-4)^2 + (y-2)^2 = 34
Thanks!
To write the equation of a circle given its center and a point on the circle, we use the formula:
(x - h)^2 + (y - k)^2 = r^2
where (h, k) represents the center of the circle, and r is the radius of the circle. In this case, the center is (4, 2), and the circle passes through (-1, -1).
Step 1: Find the radius (r)
The radius of the circle can be found using the distance formula between the center and any point on the circle. The distance formula is:
d = √[(x2 - x1)^2 + (y2 - y1)^2]
In this case, we can use the distance formula to find the distance between the center (4, 2) and the given point (-1, -1):
d = √[(-1 - 4)^2 + (-1 - 2)^2]
= √[(-5)^2 + (-3)^2]
= √[25 + 9]
= √34
Therefore, the radius (r) of the circle is √34.
Step 2: Substitute the values into the formula
Now that we have the center (h, k) = (4, 2) and the radius (r) = √34, we can substitute these values into the formula:
(x - h)^2 + (y - k)^2 = r^2
Substituting the values, we get:
(x - 4)^2 + (y - 2)^2 = (√34)^2
(x - 4)^2 + (y - 2)^2 = 34
Therefore, the equation of the circle with the given conditions is (x - 4)^2 + (y - 2)^2 = 34.