{(х-2)2+(у-3)2=16 у=х
To solve the given system of equations, let's analyze each equation step by step.
- \( (x-2)^2 + (y-3)^2 = 16 \)
- \( y = x \)
The first equation represents a circle centered at the point (2, 3) with a radius of 4 (since \( \sqrt{16} = 4 \)).
The second equation is a straight line that passes through the origin with a slope of 1 (the line \( y = x \) is at a 45-degree angle).
Step 1: Substitute Equation 2 into Equation 1
Now we substitute \( y = x \) into the first equation:
\[ (x - 2)^2 + (x - 3)^2 = 16 \]
Step 2: Expand and Simplify
Now we expand the left side:
\[ (x - 2)^2 = x^2 - 4x + 4 \]
\[ (x - 3)^2 = x^2 - 6x + 9 \]
Now substitute these expansions into the equation:
\[ (x^2 - 4x + 4) + (x^2 - 6x + 9) = 16 \]
Combine the terms:
\[ 2x^2 - 10x + 13 = 16 \]
Now, move 16 to the left side:
\[ 2x^2 - 10x + 13 - 16 = 0 \] \[ 2x^2 - 10x - 3 = 0 \]
Step 3: Apply the Quadratic Formula
To solve this quadratic equation \( ax^2 + bx + c = 0 \), where \( a = 2 \), \( b = -10 \), and \( c = -3 \), we use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Calculating the discriminant:
\[ b^2 - 4ac = (-10)^2 - 4 \cdot 2 \cdot (-3) = 100 + 24 = 124 \]
Now plug in the values into the quadratic formula:
\[ x = \frac{10 \pm \sqrt{124}}{4} \]
\[ x = \frac{10 \pm 2\sqrt{31}}{4} \] \[ x = \frac{5 \pm \sqrt{31}}{2} \]
Step 4: Find the Corresponding y-values
Since \( y = x \), the corresponding \( y \)-values are:
\[ y = \frac{5 \pm \sqrt{31}}{2} \]
Conclusion
The solutions to the system of equations are:
\[ \left( \frac{5 + \sqrt{31}}{2}, \frac{5 + \sqrt{31}}{2} \right) \] and \[ \left( \frac{5 - \sqrt{31}}{2}, \frac{5 - \sqrt{31}}{2} \right) \]
These points (x, y) are where the line intersects with the circle.
