Question 916·Medium·Math

Type $x^2+y^2=25$ and $y=x+4$ on separate Desmos lines, then click both crossings. An intersection lies on both graphs, so Desmos shows the system's solutions at about $(0.915,4.915)$ and $(-4.915,-0.915)$.

The first quadrant is the upper-right region, where $x>0$ and $y>0$. So keep $(0.915,4.915)$. In $(x,y)$, $x$ is the first coordinate, about $0.915$; $4.915$ is the $y$-coordinate.

Type $\frac{\sqrt{34}+4}{2}$ and $\frac{\sqrt{34}-4}{2}$ into Desmos. They print about $4.915$ and $0.915$, respectively. Match the first coordinate, not the second: the first-quadrant solution has $x=\frac{\sqrt{34}-4}{2}$. Choice D.

The line says $y=x+4$, so replace $y$ with the whole expression $x+4$ in the circle equation. This is substitution: $x^2+(x+4)^2=25$.

Expand the whole square, $(x+4)^2$: $x^2+x^2+8x+16=25$. Combine the $x^2$ terms: $2x^2+8x+16=25$. Subtract $25$ to put the quadratic in the form $ax^2+bx+c=0$: $2x^2+8x-9=0$.

For $ax^2+bx+c=0$, the quadratic formula is $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. Here $a=2$, $b=8$, and the constant is negative, $c=-9$. Substitute, keeping that sign: $x=\frac{-8\pm\sqrt{8^2-4(2)(-9)}}{2(2)}$.

Calculate under the square root: $x=\frac{-8\pm\sqrt{64+72}}{4}$. Add: $x=\frac{-8\pm\sqrt{136}}{4}$. Since $136=4\times34$, take $\sqrt4=2$ out of the root: $x=\frac{-8\pm2\sqrt{34}}{4}$. Divide every term by $2$: $x=\frac{-4\pm\sqrt{34}}{2}$. Keep both signs until you use the quadrant condition.

Because $\sqrt{34}>4$, only the plus sign gives $x>0$. Then $y=x+4>0$ too, so that point is in the first quadrant. Its $x$-coordinate is $\frac{\sqrt{34}-4}{2}$. Choice D.