The Equation of a Circle: Free Response

Substitute the two points into the distance formula, computing each coordinate gap first.

$d = \sqrt{(7-2)^2 + (9-(-3))^2} = \sqrt{5^2 + 12^2}.$

Square each gap and add before taking any root, since $5^2+12^2 = 25+144 = 169$:

$d = \sqrt{169} = 13.$

Substitute the center $(2,-3)$ and the radius from part A into the standard form $(x-h)^2+(y-k)^2=r^2$, taking care with the sign of a negative center coordinate.

$(x-2)^2+\bigl(y-(-3)\bigr)^2 = 13^2.$

Simplify $y-(-3)$ to $y+3$ and $13^2$ to $169$:

$(x-2)^2+(y+3)^2=169.$

The origin-centered form $x^2+y^2=r^2$ means the right-hand side is already $r^2$, so there is no need to simplify $\sqrt{200}$: compare SQUARED distances instead.

$(-10-0)^2+(10-0)^2 = 100+100 = 200.$

This squared distance equals the right-hand side of the circle's equation exactly, so the point sits precisely $r$ from the origin, neither closer nor farther:

$200 = 200 \quad\Longrightarrow\quad \text{the point lies ON the circle.}$

Check Marcus's claim against the standard form $(x-h)^2+(y-k)^2=r^2$ piece by piece, rather than assuming everything he wrote is wrong.

The first parenthesis is $x+3$, which is $x-(-3)$, so $h=-3$, not the $3$ Marcus wrote:

$x+3 = x-(-3) \quad\Longrightarrow\quad h=-3.$

The second parenthesis is $y-4$, which already matches the form $y-k$ directly, so $k=4$ really is correct.

The right-hand side is $r^2=20$, not $r$ itself, so the radius is its square root, simplified:

$r=\sqrt{20}=\sqrt{4\cdot5}=2\sqrt5.$

So Marcus's center is wrong in its $x$-coordinate only, and his radius is wrong because he never took the square root.

Substitute the CORRECTED center into the distance formula, not Marcus's claimed one.

$d=\sqrt{\bigl(-1-(-3)\bigr)^2+(8-4)^2}=\sqrt{2^2+4^2}.$

Square, add, and simplify the root:

$d=\sqrt{4+16}=\sqrt{20}=2\sqrt5.$

This matches the corrected radius exactly, so the point sits precisely on the circle rather than merely near it.

Marcus's rule is a generalization from a single glance, so test it against the form itself rather than against one example.

The standard form is built from subtraction:

$(x-h)^2+(y-k)^2=r^2.$

Whenever the parentheses show a plus sign, such as $x+3$, that plus sign can only appear because the SUBTRACTED value $h$ is itself negative:

$x+3 = x-(-3).$

So a plus sign is not a positive coordinate in disguise, it is the form's way of subtracting a negative one. Marcus's rule has the relationship exactly backwards.

The center of a circle determined by a diameter is the midpoint of its two endpoints, found by averaging each coordinate separately.

$\left(\frac{-4+2}{2},\ \frac{-1+3}{2}\right) = \left(\frac{-2}{2},\ \frac{2}{2}\right) = (-1,\,1).$

The radius is the distance from the center to either endpoint, and only $r^2$ is needed for the standard form, so square and add without taking any root.

$r^2=\bigl(2-(-1)\bigr)^2+(3-1)^2=3^2+2^2=9+4=13.$

Substitute the center and this $r^2$ into $(x-h)^2+(y-k)^2=r^2$, watching the sign of the negative center coordinate:

$(x-(-1))^2+(y-1)^2=13 \quad\Longrightarrow\quad (x+1)^2+(y-1)^2=13.$

Compare the SQUARED distance from the center to the proposed point against $r^2$ directly, the same shortcut used in part B.

$\bigl(2-(-1)\bigr)^2+(1-1)^2=3^2+0^2=9.$

Since this squared distance is less than $r^2$, the proposed point sits closer to the center than the border does, so it lies inside the garden bed.

Both squared terms share a coefficient of $3$, so divide EVERY term of the equation by $3$, including the constant.

$\frac{3x^2}{3}+\frac{3y^2}{3}+\frac{24x}{3}-\frac{30y}{3}+\frac{15}{3}=0.$

Simplify each term:

$x^2+y^2+8x-10y+5=0.$

Group the $x$-terms and $y$-terms, and move the lone constant to the other side.

$(x^2+8x)+(y^2-10y)=-5.$

Complete the square on each group separately: half of $8$ is $4$, squared is $16$; half of $-10$ is $-5$, squared is $25$. Add both constants to BOTH sides.

$(x^2+8x+16)+(y^2-10y+25)=-5+16+25.$

Each group is now a perfect square, and the right side totals $36$:

$(x+4)^2+(y-5)^2=36.$

Read the center and radius directly off the standard form, watching the sign convention.

$(x-(-4))^2+(y-5)^2=6^2 \quad\Longrightarrow\quad \text{center }(-4,5),\ \text{radius }6.$

Substitute the point into the standard-form equation to check it, rather than estimating its position:

$(2-(-4))^2+(5-5)^2 = 6^2+0^2=36.$

This matches the right-hand side exactly, so the point lies precisely on the circle.

Group the $x$-terms and $y$-terms, and move the constant to the other side.

$(x^2-6x)+(y^2+10y)=-34.$

Complete the square on each group: half of $-6$ is $-3$, squared is $9$; half of $10$ is $5$, squared is $25$. Add both constants to both sides.

$(x^2-6x+9)+(y^2+10y+25)=-34+9+25.$

Simplify the right side:

$(x-3)^2+(y+5)^2=0.$

A sum of two real squares can equal $0$ only when EACH square is $0$, which forces $x=3$ and $y=-5$. The graph is not a curve at all, just the single point $(3,-5)$.

Group the $x$-terms and $y$-terms, and move the constant to the other side.

$(x^2+2x)+(y^2-4y)=-9.$

Complete the square on each group: half of $2$ is $1$, squared is $1$; half of $-4$ is $-2$, squared is $4$. Add both constants to both sides.

$(x^2+2x+1)+(y^2-4y+4)=-9+1+4.$

Simplify the right side:

$(x+1)^2+(y-2)^2=-4.$

A sum of two real squares can never be negative, so no real point $(x,y)$ can make the left side equal $-4$. The graph is empty.

The left side, $(x-h)^2+(y-k)^2$, is a sum of two real squares, and a real square is never negative, so their sum is never negative either:

$(x-h)^2 \ge 0 \quad\text{and}\quad (y-k)^2 \ge 0 \quad\Longrightarrow\quad (x-h)^2+(y-k)^2 \ge 0.$

That single fact governs all three cases.

If $R>0$, the sum of two nonnegative squares can equal a positive number in many ways, tracing out an honest curve: a circle of radius $\sqrt{R}$.

If $R=0$, the sum can equal $0$ in exactly one way, both squares individually $0$, which pins down a single point rather than a curve.

If $R<0$, the sum can never equal it at all, since a sum of nonnegative numbers cannot be negative, so no real point satisfies the equation and the graph is empty.

Parts A and B are exactly the last two cases: one right side came out to $0$ and produced a single point, and the other came out negative and produced no graph. Only a POSITIVE right side, $R>0$, produces a genuine circle.