Describe and sketch the region in the plane defined by the simultaneous inequalities x^2 + y^2 <= 9 and y <= x + 1, including the points where the boundary line meets the circle.

Step-by-step solution

1. 1

   Interpret the first inequality as a filled disk. The equation $x^{2}+y^{2}=9$ is the circle centred at the origin with radius $\sqrt{9}=3$. Because the inequality is $\leq$, the region is the circle together with its interior — a closed disk. Draw the circle as a solid curve to signal that the boundary is included.

2. 2

   Interpret the second inequality as a closed half-plane. The boundary $y=x+1$ is a line of slope $1$ through $(0,1)$ and $(-1,0)$. Since $y\leq x+1$, we keep the side below the line; test the origin, where $0\leq 0+1$ is true, so the half-plane containing $(0,0)$ is the correct one. Draw this line solid too.

3. 3

   Find where the boundary line cuts the circle. Substitute $y=x+1$ into $x^{2}+y^{2}=9$:

   $x^{2}+(x+1)^{2}=9\ \Longrightarrow\ 2x^{2}+2x+1=9\ \Longrightarrow\ 2x^{2}+2x-8=0\ \Longrightarrow\ x^{2}+x-4=0$

   The quadratic formula gives

   $x=\frac{-1\pm\sqrt{1+16}}{2}=\frac{-1\pm\sqrt{17}}{2}$

4. 4

   Convert to the two intersection points. Using $y=x+1$ on each root,

   $\left(\frac{-1+\sqrt{17}}{2},\ \frac{1+\sqrt{17}}{2}\right)\quad\text{and}\quad\left(\frac{-1-\sqrt{17}}{2},\ \frac{1-\sqrt{17}}{2}\right)$

   Numerically these are about $(1.5616,\,2.5616)$ and $(-2.5616,\,-1.5616)$; substituting either into $x^{2}+y^{2}$ returns exactly $9.0$, so both really do lie on the circle. Since the discriminant $17>0$, the line is a genuine secant, not a tangent.

5. 5

   Describe the resulting region. The answer is the major portion of the closed disk: everything inside or on the circle of radius $3$ that also lies on or below the line $y=x+1$. The chord joining the two intersection points slices off the small circular segment above the line, and that segment is the only part of the disk excluded. Both boundary pieces — the circular arc and the chord — belong to the region because both inequalities are non-strict.