Question

1 multiple choice 1 point for what values of r does the line y = x - 4 intersect (at one or more points) the circle x² + y² = r²? r ≥ 2√2 r ≥ 1/2 r ≥ 4√2 r ≥ 1 2 multiple choice 1 point what is the solution of the following system of equations? y = (x + 3)² - 4 y = 2x + 5 (-4, -3) and (0, 5) (-4, 0) (0, -4) (-3, -4) and (5, 0) 3 multiple choice 1 point if x + 2y = 12 and x - y = 3 find the value of x + y. none of these 3 5 9

Explanation:

Step1: Substitute $y$ in circle equation

Substitute $y=x - 4$ into $x^{2}+y^{2}=r^{2}$. We get $x^{2}+(x - 4)^{2}=r^{2}$. Expand $(x - 4)^{2}$: $x^{2}+x^{2}-8x + 16=r^{2}$, which simplifies to $2x^{2}-8x+16 - r^{2}=0$.

Step2: Use discriminant formula

For a quadratic equation $ax^{2}+bx + c = 0$ ($a = 2$, $b=-8$, $c = 16 - r^{2}$), the discriminant $\Delta=b^{2}-4ac$. Here, $\Delta=(-8)^{2}-4\times2\times(16 - r^{2})$. Calculate $\Delta = 64-128 + 8r^{2}=8r^{2}-64$.

Step3: Set $\Delta\geq0$ for intersection

Since the line intersects the circle at one or more points, $\Delta\geq0$. So, $8r^{2}-64\geq0$. Add 64 to both sides: $8r^{2}\geq64$. Divide both sides by 8: $r^{2}\geq8$. Take the square - root of both sides, considering $r\gt0$ (radius), we have $r\geq2\sqrt{2}$.

Step1: Set the two equations equal

Set $(x + 3)^{2}-4=2x + 5$. Expand $(x + 3)^{2}$: $x^{2}+6x+9-4=2x + 5$. Simplify to get $x^{2}+6x + 5=2x + 5$.

Step2: Rearrange to quadratic form

Subtract $2x$ and 5 from both sides: $x^{2}+4x=0$. Factor out $x$: $x(x + 4)=0$.

Step3: Solve for $x$

Set each factor equal to zero. If $x=0$, then $y=2\times0 + 5=5$. If $x=-4$, then $y=2\times(-4)+5=-3$.

Step1: Solve the system of equations

We have the system

$$\begin{cases}x + 2y=12\\x - y=3\end{cases}$$

. Subtract the second equation from the first: $(x + 2y)-(x - y)=12 - 3$. This gives $3y=9$, so $y = 3$.

Step2: Find the value of $x$

Substitute $y = 3$ into $x - y=3$, we get $x-3=3$, so $x=6$.

Step3: Calculate $x + y$

$x + y=6 + 3=9$.

Answer:

$r\geq2\sqrt{2}$