Write the equation of the circle centered at (6,10) that passes through (-9,1)
centre at (6,10), so the equation is
(x-6)^2 + (y-10)^2 = r^2
sub in your point (-9,1) to solve for r^2
then rewrite my equation. All done!
To find the equation of a circle, you need two key pieces of information: the center coordinates and either the radius or any point on the circle. In this case, you have the center coordinates, (6,10), and a point on the circle, (-9,1).
The general equation of a circle with center coordinates (h, k) and radius r is:
(x - h)^2 + (y - k)^2 = r^2
Given that the center is at (6, 10), let's substitute the values into the equation:
(x - 6)^2 + (y - 10)^2 = r^2
Now we need to find the radius. We can use the distance formula, which calculates the distance between two points, to find the radius. The distance formula is:
d = √[(x2 - x1)^2 + (y2 - y1)^2]
In this case, we can use the points (-9,1) and (6,10) to calculate the radius:
r = √[(-9 - 6)^2 + (1 - 10)^2]
= √[(-15)^2 + (-9)^2]
= √[225 + 81]
= √306
Now, substitute the radius into the equation:
(x - 6)^2 + (y - 10)^2 = (√306)^2
Simplifying:
(x - 6)^2 + (y - 10)^2 = 306
Therefore, the equation of the circle centered at (6,10) that passes through (-9,1) is (x - 6)^2 + (y - 10)^2 = 306.