how do i write this equation...
circle with center (1,-3) that passes through the point (4,2).
I have (x-1)^2 + (y+3)^2=? how do i get the radius...
complete your equation to
(x-1)^2 + (y+3)^2= r^2
but (4,2) lies on your circle, so ...
(4-1)^2 + (2+3)^2 = r^2
etc.
thanks:)
To find the equation of a circle, you need to know the center and the radius. In this case, you already have the center, which is (1, -3). To find the radius, you can use the distance formula between the center and the given point (4, 2).
The distance formula between two points (x1, y1) and (x2, y2) is:
√((x2 - x1)^2 + (y2 - y1)^2)
Let's calculate the distance between the center (1, -3) and the point (4, 2):
√((4 - 1)^2 + (2 - (-3))^2)
= √(3^2 + 5^2)
= √(9 + 25)
= √34
So, the radius of the circle is √34.
Now, you can write the equation of the circle using the center and the radius. The general equation of a circle is:
(x - h)^2 + (y - k)^2 = r^2
Where (h, k) represents the center and r represents the radius. Plugging in the values we have:
(x - 1)^2 + (y + 3)^2 = (√34)^2
Simplifying, we have:
(x - 1)^2 + (y + 3)^2 = 34
Thus, the equation of the circle with center (1, -3) that passes through the point (4, 2) is (x - 1)^2 + (y + 3)^2 = 34.