SAT Question e833de9a: Nonlinear Equations In One Variable And Systems Of Equations In Two Variables answer and explanation with Desmos solution

Choice D is correct. It’s given that the graphs of $x^2+y^2=36$ and $y=mx+\frac{b}{4}$, where $m$ and $b$ are negative constants, intersect at the point $(-5,y)$ in the xy-plane. It follows that this intersection point represents a solution to the given system of equations. Substituting $-5$ for $x$ in the equation $x^2+y^2=36$ yields $(-5{)}^2+y^2=36$, or $25+y^2=36$. Subtracting $25$ from each side of this equation yields $y^2=11$. It's also given that $y<0$, so taking the negative square root of each side of this equation yields $y=-\sqrt{11}$. Substituting $-5$ for $x$ and $-\sqrt{11}$ for $y$ in the equation $y=mx+\frac{b}{4}$ yields $-\sqrt{11}=m(-5)+\frac{b}{4}$, or $-\sqrt{11}=-5m+\frac{b}{4}$. Adding $5m$ to each side of this equation yields $5m-\sqrt{11}=\frac{b}{4}$. Multiplying each side of this equation by $4$ yields $4(5m-\sqrt{11})=b$. Applying the distributive property to the left-hand side of this equation yields $20m-4\sqrt{11}=b$. Therefore, the expression $20m-4\sqrt{11}$ represents the value of $b$.