Write the equation of the circle with center (-2,-3) and tangent to the line 2x+5y=10.

let P(a,b) be the point on the circle where the tangent hits it.

slope of the given tangent is -2/5
so the slope of the radius to P is +5/2

but slope of radius is (b+3)/(a+2)
so (b+3)/(a+2) = 5/2
5a - 2a = -4 (#1)

but P(a,b) also lies on the tangent, so
2a + 5b = 10 (#2)

taking 5(#1) + 2(#2) and adding gave me a=0
subbing that back into #2 gave be b=2

So the point of tangency is (0,2)
which gives me a radius of √29 using the distance formula between P and the centre.

so now you have the centre and the radius.....
(x+2)^2 + (y+3)^2 = 29

To find the equation of the circle with a given center and tangent to a given line, we can follow these steps:

1. Determine the radius of the circle.
- The radius of the circle is equal to the distance between the center of the circle and the line.
- We use the formula to find the distance between a point (x₁, y₁) and a line Ax + By + C = 0, which is given by: d = |Ax₁ + By₁ + C| / √(A² + B²).
- In this case, the center of the circle is (-2, -3) and the line is 2x + 5y = 10. So, A = 2, B = 5, C = -10, and (x₁, y₁) = (-2, -3).
- Plug these values into the formula to find the radius.
d = |2(-2) + 5(-3) + (-10)| / √(2² + 5²)
= |-4 - 15 - 10| / √(4 + 25)
= |-29| / √29
= 29 / √29
= √29

2. Use the formula for the equation of a circle, which is:
(x - h)² + (y - k)² = r²
where (h, k) is the center of the circle and r is the radius.

Plug in the values we have:
(x - (-2))² + (y - (-3))² = (√29)²
(x + 2)² + (y + 3)² = 29

Therefore, the equation of the circle with center (-2, -3) and tangent to the line 2x + 5y = 10 is (x + 2)² + (y + 3)² = 29.