Find the equation of a circle passing through the origin and the points where the line 3x + 4y=12 meets the axes of the co-ordinates..
the circle passes through
(0,0), (4,0), (0,3)
If the center is (h,k) and the radius is r, then
h^2 + k^2 = r^2
(4-h)^2 + k^2 = r^2
h^2 + (3-k)^2 = r^2
The circle is thus
(x-2)^2 + (y-3/2)^2 = (5/2)^2
see
http://www.wolframalpha.com/input/?i=%28x-2%29^2+%2B+%28y-3%2F2%29^2+%3D+%285%2F2%29^2
How can we say that the circle passes through (4,0),(0,3)???
To find the equation of a circle passing through the origin and the points where the line 3x + 4y = 12 meets the axes, we can follow these steps:
1. Find the x-intercept of the given line:
To find the x-intercept, we set y = 0 and solve for x:
3x + 4(0) = 12
3x = 12
x = 4
So, the x-intercept is (4, 0).
2. Find the y-intercept of the given line:
To find the y-intercept, we set x = 0 and solve for y:
3(0) + 4y = 12
4y = 12
y = 3
So, the y-intercept is (0, 3).
3. Find the radius of the circle:
The radius of the circle is the distance between the origin (0, 0) and one of the intercept points. We can use the distance formula:
d = √((x2 - x1)^2 + (y2 - y1)^2)
Let's calculate the distance from the origin to the x-intercept:
d = √((4 - 0)^2 + (0 - 0)^2)
d = √(16 + 0)
d = √16
d = 4
So, the radius of the circle is 4.
4. Find the center of the circle:
Since the origin (0, 0) is the center of the circle and the radius is 4 units, the equation of the circle is:
(x - 0)^2 + (y - 0)^2 = 4^2
Simplifying further, we get:
x^2 + y^2 = 16
Therefore, the equation of the circle passing through the origin and the points where the line 3x + 4y = 12 meets the axes is x^2 + y^2 = 16.