How to Find the X-Intercept of a Circle
- 1). Plug the values of h, k, and r into the equation. So, if their values are 2, 4, and 6, respectively, then the equation becomes (X - 2)^2 + (Y - 4)^2 = 6^2. When simplified, it turns into (X - 2)^2 + (Y - 4)^2 = 36.
- 2). Set the value of y to 0. Whenever a circle is touching the x-axis, the value of y must be 0. The equation then becomes (X - 2)^2 + (0 - 4)^2 = 36. When simplified, it turns into (X - 2)^2 + (-4)^2 = 36, or (X - 2)^2 + 16 = 36.
- 3). Simplify the equation as much as possible. Using the FOIL method (multiply the First terms, multiply the Outer terms, multiply the Inner terms, multiple the Last term), we can turn the equation (X-2)^2 + 16 = 36 into (x^2 -4x + 4) + 16 = 36. Then, we subtract 36 from both sides to get X^2 - 4x + 4 + 16 - 36 = 0, or x^2 -4x -16.
- 4). Solve the equation by using the quadratic formula, which states that x = (-b +/- the square root of b^2 - 4ac) divided by 2a. This is based on the premise that ax^2+bx+c=0. So in the case of X^2 - 4x - 16 = 0, a = 1, b = -4, and c = -16. Using the quadratic formula, we discover two values for x: 6.5 and -2.5.
- 5). Determine the x-intercept by using the values obtained from the quadratic formula. In our case, the x-intercept consists of two points, including (6.5, 0) and (-2.5, 0).