how would you find an equation for a circle that passes through the origin and has intercepts equal to 1 and 2 on the x and y axes
The general equation for a circle is:
(x-a)^2 + (y-b)^2 = R^2
R is the radius of the circle and the point (a,b) is the center. To solve the problem you just insert the coordinates of the three points that are known to be on the circle in the equation and solve for a, b and R:
Origin: x = 0, y = 0:
a^2 + b^2 = R^2
x= 1, y= 0:
(1-a)^2 +b^2 = R^2
x= 0, y= 2:
a^2 + (2-b)^2 = R^2
Subtract from the last two equations the first equation to obtain two equations one containing only a and the other containing only b.
ok so i found:
(y-1)^2 + (x-.5)^2 - r^2
so how do you find r??
To find the equation of a circle that passes through the origin and has intercepts equal to 1 and 2 on the x and y axes, you can follow these steps:
1. Start with the standard equation of a circle: (x - h)^2 + (y - k)^2 = r^2, where (h, k) represents the center of the circle and r represents the radius.
2. Since the circle passes through the origin, the center (h, k) must be (0, 0).
3. Substitute the values of h and k into the equation to get x^2 + y^2 = r^2.
4. The intercepts on the x and y axes are given as 1 and 2, respectively. The x-intercept is the point (1, 0), and the y-intercept is the point (0, 2).
5. Use these points to find the radius (r).
- Substitute (1, 0) into the equation, which gives you 1^2 + 0^2 = r^2. This simplifies to r^2 = 1^2 = 1.
- Substitute (0, 2) into the equation, which gives you 0^2 + 2^2 = r^2. This simplifies to r^2 = 2^2 = 4.
- Since (h, k) is the origin, the radius (r) is the distance from the origin to either of the intercept points. Using the Pythagorean theorem, r = √(1^2 + 2^2) = √5.
6. Substitute the value of r into the equation x^2 + y^2 = r^2, which gives you x^2 + y^2 = (√5)^2. Simplifying, you get x^2 + y^2 = 5.
Therefore, the equation of the circle that passes through the origin and has intercepts equal to 1 and 2 on the x and y axes is x^2 + y^2 = 5.