Write the standard equation for the circle center (-6,-8), that passes through (0,0).
I know that the formula would leave the 1qstr part as
(x+6)+(y+8)
I'm not sure about the radius when it goes through (0,0)
To find the radius of the circle that passes through the point (0,0), we can use the distance formula. The distance between the center (-6,-8) and (0,0) will give us the radius.
The distance formula is given by:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Plugging in the coordinates, we have:
radius = sqrt((0 - (-6))^2 + (0 - (-8))^2)
radius = sqrt(6^2 + 8^2)
radius = sqrt(36 + 64)
radius = sqrt(100)
radius = 10
Therefore, the radius of the circle is 10 units.
The standard equation for a circle is:
(x - h)^2 + (y - k)^2 = r^2
- where (h, k) represents the center coordinates and r represents the radius.
Plugging in the given values, we have:
(x - (-6))^2 + (y - (-8))^2 = 10^2
(x + 6)^2 + (y + 8)^2 = 100
Hence, the standard equation for the circle with center (-6,-8) and passing through (0,0) is (x + 6)^2 + (y + 8)^2 = 100.
To find the standard equation for a circle, we need to know the center and the radius. Since we are given the center as (-6,-8) and the circle passes through (0,0), we can use the distance formula to find the radius.
The distance formula between two points (x1, y1) and (x2, y2) is given by:
distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
In this case, substituting the given coordinates into the distance formula, we have:
radius = sqrt((0 - (-6))^2 + (0 - (-8))^2)
= sqrt((6)^2 + (8)^2)
= sqrt(36 + 64)
= sqrt(100)
= 10
Now that we have the radius, we can write the standard equation for the circle as:
(x - h)^2 + (y - k)^2 = r^2
Substituting the given values into the equation, we get:
(x - (-6))^2 + (y - (-8))^2 = 10^2
(x + 6)^2 + (y + 8)^2 = 100
Therefore, the standard equation for the circle with center (-6,-8) that passes through (0,0) is (x + 6)^2 + (y + 8)^2 = 100.
the distance from (-6,-8) to (0,0) is 10. So,
(x+6)^2 + (y+8)^2 = 10^2