The equations provided are:

1. $(x - 4)^2 + (y - 3)^2 = 98$
2. $-2x + y = 2$

Step 1: Identify the Equations' Forms

1. The first equation represents a circle with center $(4, 3)$ and radius $\sqrt{98}$.
2. The second equation is a linear equation that we can solve for $y$ in terms of $x$.

Step 2: Solve for $y$ in Terms of $x$

From the second equation: $y = 2 + 2x$

Step 3: Substitute $y$ in the Circle Equation

Now, substitute $y = 2 + 2x$ into the circle equation: $(x - 4)^2 + ((2 + 2x) - 3)^2 = 98$ Simplify $(2 + 2x - 3)$ to obtain: $(x - 4)^2 + (2x - 1)^2 = 98$

Step 4: Expand and Solve the Equation

1. Expand $(x - 4)^2$ and $(2x - 1)^2$: $(x - 4)^2 = x^2 - 8x + 16$ $(2x - 1)^2 = 4x^2 - 4x + 1$

2. Substitute these expansions back into the equation: $x^2 - 8x + 16 + 4x^2 - 4x + 1 = 98$

3. Combine like terms: $5x^2 - 12x + 17 = 98$

4. Move all terms to one side to set the equation to zero: $5x^2 - 12x - 81 = 0$

Step 5: Solve the Quadratic Equation

Now we solve $5x^2 - 12x - 81 = 0$ using the quadratic formula: $x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4 \cdot 5 \cdot (-81)}}{2 \cdot 5}$ $x = \frac{12 \pm \sqrt{144 + 1620}}{10}$ $x = \frac{12 \pm \sqrt{1764}}{10}$ $x = \frac{12 \pm 42}{10}$

This gives us two solutions for $x$:

1. $x = \frac{54}{10} = 5.4$
2. $x = \frac{-30}{10} = -3$

Step 6: Find Corresponding $y$ Values

For $x = 5.4$: $y = 2 + 2(5.4) = 2 + 10.8 = 12.8$

For $x = -3$: $y = 2 + 2(-3) = 2 - 6 = -4$

Solution

The points of intersection are:

1. $(5.4, 12.8)$
2. $(-3, -4)$

Would you like further details on any of these steps?

Here are some related questions you might be interested in:

1. How can we determine the intersection points graphically?
2. What other methods can be used to solve the intersection of a circle and a line?
3. Can we verify the points of intersection using substitution in the original equations?
4. How do we determine if a line is tangent to a circle?
5. What does it mean if there are no real solutions in the quadratic step?

Tip: When dealing with circles and lines, translating one equation to express one variable in terms of another and substituting can simplify finding intersections.