Question 47·SAT Desmos Regression Practice·Geometry and Trigonometry

Every point on the circle satisfies the given equation. Move $ax+by+c$ to the right so Desmos can fit those unknown coefficients: $x^2+y^2=-ax-by-c$.Call the coefficients on the right $u$, $v$, and $w$. Moving the constant across the equals sign flips its sign, so $w=-c$.

Enter the three points as matching rows of a Desmos table. Then type $x_1^2+y_1^2\sim ux_1+vy_1+w$; the $\sim$ tells Desmos to fit the same equation to every row. Under PARAMETERS, Desmos shows $u=10$, $v=-8$, and $w=-16$.

The fit gave $w=-16$, but $w=-c$. Type $-w$ below the regression; Desmos prints $16$. So the constant in the requested equation is $c=16$. Grid in 16.

Find the slopes, or vertical changes divided by horizontal changes, from $(1,-1)$ to the other two points. Type $\frac{0-(-1)}{8-1}$ and $\frac{-8-(-1)}{2-1}$. Desmos gives $\frac17$ (use its fraction button if needed) and $-7$. These are negative reciprocals, so the segments meet at a right angle.

The right angle is at a point on the circle, so the side across from it, joining $(8,0)$ and $(2,-8)$, is a diameter: it passes through the center. Type $A=(8,0)$, $B=(2,-8)$, and $\operatorname{midpoint}(A,B)$ in Desmos. Click the plotted midpoint to read the center $(5,-4)$.

A radius runs from the center $(5,-4)$ to any point on the circle. The Pythagorean theorem says the squared distance equals the sum of the squared horizontal and vertical changes. Use $(8,0)$: type $(8-5)^2+(0-(-4))^2$. Desmos prints $25$, the squared radius $r^2$.

The standard form of a circle with center $(h,k)$ is $(x-h)^2+(y-k)^2=r^2$. Subtract $r^2$ to put it in the requested form: $(x-h)^2+(y-k)^2-r^2=0$.The terms without $x$ or $y$ are $h^2+k^2-r^2$, so $c=h^2+k^2-r^2$.

Type $5^2+(-4)^2-25$ in Desmos. It prints $16$, the constant $c$ in the circle’s equation. Grid in 16.