Question

find both pts of intersection of a chord and a circle

Answer 1

The question is general and requires specific equations for the circle and the chord to provide a numerical answer. For the example case where the circle is $x^2 + y^2 = 16$ and the chord is $x+y=2$, the points of intersection are $(1+\sqrt{7}, 1-\sqrt{7})$ and $(1-\sqrt{7}, 1+\sqrt{7})$.

Answer 2

1. Represent Equations: Write down the equation of the circle and the equation of the chord (which is a straight line).
2. Substitution: From the linear equation of the chord, express one variable (e.g., $y$) in terms of the other (e.g., $x$).
3. Form Quadratic Equation: Substitute this expression into the circle's equation. This will result in a quadratic equation in a single variable.
4. Solve Quadratic: Solve the quadratic equation to find the two possible values for that variable.
5. Find Corresponding Values: Substitute these values back into the chord's linear equation to find the corresponding values of the other variable.
6. Intersection Points: The pairs of coordinates $(x, y)$ obtained are the two points of intersection.

Example: Let's find the points of intersection of the circle $x^2 + y^2 = 16$ and the chord given by the line $x + y = 2$.

1. Equations:
   • Circle: $x^2 + y^2 = 16$
   • Chord: $x + y = 2$
2. Substitution: From the chord equation, $y = 2 - x$.
3. Form Quadratic Equation: Substitute $y = 2 - x$ into the circle's equation: $x^2 + (2 - x)^2 = 16$ $x^2 + (4 - 4x + x^2) = 16$ $2x^2 - 4x + 4 = 16$ $2x^2 - 4x - 12 = 0$ $x^2 - 2x - 6 = 0$
4. Solve Quadratic: Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-6)}}{2(1)} = \frac{2 \pm \sqrt{4 + 24}}{2} = \frac{2 \pm \sqrt{28}}{2} = \frac{2 \pm 2\sqrt{7}}{2} = 1 \pm \sqrt{7}$ So, $x_1 = 1 + \sqrt{7}$ and $x_2 = 1 - \sqrt{7}$.
5. Find Corresponding Values: Substitute these $x$ values back into $y = 2 - x$:
   • For $x_1 = 1 + \sqrt{7}$: $y_1 = 2 - (1 + \sqrt{7}) = 1 - \sqrt{7}$
   • For $x_2 = 1 - \sqrt{7}$: $y_2 = 2 - (1 - \sqrt{7}) = 1 + \sqrt{7}$
6. Intersection Points: The points are $(1 + \sqrt{7}, 1 - \sqrt{7})$ and $(1 - \sqrt{7}, 1 + \sqrt{7})$.