a cricle with centre at (0,5) passes through (-6,-3) help me find the equation in the standard from

Sure, let's find the equation of the circle.

The general equation of a circle is (x-h)² + (y-k)² = r², where (h,k) is the center of the circle and r is the radius.

Here, the center of the circle is given as (0,5). So, h = 0 and k = 5.

The circle passes through the point (-6,-3). We can use this point to find the radius of the circle. The radius is the distance between the center of the circle and any point on the circle. We can use the distance formula to find this:

r = sqrt[(x-h)² + (y-k)²] r = sqrt[(-6-0)² + (-3-5)²] r = sqrt[(-6)² + (-8)²] r = sqrt[36 + 64] r = sqrt[100] r = 10

Now we can write the equation of the circle:

(x-0)² + (y-5)² = 10² x² + (y-5)² = 100

So, the equation of the circle in standard form is x² + (y-5)² = 100.