{(х-2)2+(у-3)2=16 {у=х

You've provided two mathematical equations:

1. \((x - 2)^2 + (y - 3)^2 = 16\) (which represents a circle centered at (2, 3) with a radius of 4)
2. \(y = x\) (which represents a line at a 45-degree angle through the origin)

To find the points of intersection between the circle and the line, we can substitute \(y = x\) into the circle's equation.

Starting with the circle's equation:

\[ (x - 2)^2 + (y - 3)^2 = 16 \]

Substituting \(y = x\):

\[ (x - 2)^2 + (x - 3)^2 = 16 \]

Now, expand the squares:

\[ (x - 2)^2 = x^2 - 4x + 4 \] \[ (x - 3)^2 = x^2 - 6x + 9 \]

Combining these, we have:

\[ (x^2 - 4x + 4) + (x^2 - 6x + 9) = 16 \] \[ 2x^2 - 10x + 13 = 16 \]

Now, let's set it to zero:

\[ 2x^2 - 10x + 13 - 16 = 0 \] \[ 2x^2 - 10x - 3 = 0 \]

Dividing the entire equation by 2 to simplify:

\[ x^2 - 5x - \frac{3}{2} = 0 \]

Next, we can use the quadratic formula to find the values of \(x\):

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Where \(a = 1\), \(b = -5\), and \(c = -\frac{3}{2}\):

\[ x = \frac{5 \pm \sqrt{(-5)^2 - 4(1)(-\frac{3}{2})}}{2(1)} \] \[ x = \frac{5 \pm \sqrt{25 + 6}}{2} \] \[ x = \frac{5 \pm \sqrt{31}}{2} \]

Now, we have two values for \(x\):

1. \(x_1 = \frac{5 + \sqrt{31}}{2}\)
2. \(x_2 = \frac{5 - \sqrt{31}}{2}\)

Next, we can find the corresponding \(y\)-values using \(y = x\):

1. \(y_1 = x_1 = \frac{5 + \sqrt{31}}{2}\)
2. \(y_2 = x_2 = \frac{5 - \sqrt{31}}{2}\)

Therefore, the points of intersection are:

1. \(\left(\frac{5 + \sqrt{31}}{2}, \frac{5 + \sqrt{31}}{2}\right)\)
2. \(\left(\frac{5 - \sqrt{31}}{2}, \frac{5 - \sqrt{31}}{2}\right)\)

These are the intersection points of the given circle and the line.