Circles: Free Response

The distance from the center $(-6, 4)$ to a point $(x, y)$ on the boundary equals the range, $9$, so square that condition directly rather than working from a memorized template.

$(x-(-6))^2+(y-4)^2=9^2 \;\Longrightarrow\; (x+6)^2+(y-4)^2=81.$

Expand each square separately and collect the constants onto one side.

$x^2+12x+36+y^2-8y+16=81 \;\Longrightarrow\; x^2+y^2+12x-8y+52=81.$

Move the $52$ across to finish in general form.

$x^2+y^2+12x-8y-29=0.$

Substitute $x=0$ and $y=0$ into the left side of the general-form equation, rather than converting back to center-radius form first.

$0^2+0^2+12(0)-8(0)-29=-29.$

This value is negative, which means the point sits on the side of the boundary where the left side has not yet reached $0$. Because the same expression equals $(x-h)^2+(y-k)^2-r^2$ by construction, a negative value means the point's squared distance from the center is less than $r^2$, so the hiker is inside the range.

$-29<0 \;\Longrightarrow\; \text{inside the range.}$

Part A's derivation is the key: starting from $(x-h)^2+(y-k)^2=r^2$ and expanding produces exactly $x^2+y^2+Dx+Ey+F=0$, with $D=-2h$, $E=-2k$, $F=h^2+k^2-r^2$. Reversing that expansion for any point $(x,y)$, not just one on the boundary, shows the general-form expression IS the difference between squared distance and $r^2$.

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

So evaluating the left side at a point and comparing to $0$ is the same comparison, term for term, as comparing the point's squared distance from the center to $r^2$: both are negative together, zero together, and positive together. No separate calculation is needed.

Solve the line for $y$ first: $y=-x+15$, so $y+2=-x+17$. Substitute into the circle's equation.

$(x-3)^2+(-x+17)^2=100.$

Expand both squares and collect like terms.

$x^2-6x+9+x^2-34x+289=100 \;\Longrightarrow\; 2x^2-40x+198=0.$

Divide by $2$ and compute the discriminant of $x^2-20x+99=0$.

$(-20)^2-4(1)(99)=400-396=4.$

The discriminant is positive, so the quadratic has two distinct real roots and the footpath is a secant, crossing the hazard boundary at two points.

Solve $x^2-20x+99=0$ by factoring or the quadratic formula: it factors as $(x-9)(x-11)=0$, so $x=9$ or $x=11$.

A value of $x$ alone is not a point. Substitute each root back into the line $y=-x+15$.

$x=9 \Rightarrow y=6, \qquad x=11 \Rightarrow y=4.$

The crossing points are $(9,6)$ and $(11,4)$.

Test the claim directly by finding the midpoint of the two crossing points and comparing it to the site office.

$\left(\frac{9+11}{2}, \frac{6+4}{2}\right)=(10, 5).$

This midpoint is $(10,5)$, not the office at $(3,-2)$, so the two crossing points are not symmetric about the office. Symmetry about the center only happens when the chord passes through the center itself, meaning the line is a diameter of the hazard circle; this particular footpath does not pass through $(3,-2)$, so there is no reason to expect that symmetry here.

Since $y=2x+c$, we have $y+1=2x+c+1$. Substitute into the circle's equation.

$(x-1)^2+(2x+c+1)^2=20.$

Expand both squares.

$x^2-2x+1+4x^2+4x(c+1)+(c+1)^2=20.$

Collect like powers of $x$ and move the $20$ across.

$5x^2+(4c+2)x+\big[(c+1)^2-19\big]=0.$

Compute the discriminant of $5x^2+(4c+2)x+\big[(c+1)^2-19\big]=0$ and set it to $0$.

$(4c+2)^2-4(5)\big[(c+1)^2-19\big]=0.$

Expand both pieces.

$16c^2+16c+4-20(c^2+2c+1)+380=0.$

Collect like terms.

$-4c^2-24c+364=0.$

Divide by $-4$ and solve the resulting quadratic in $c$.

$c^2+6c-91=0 \;\Longrightarrow\; c=\frac{-6\pm\sqrt{400}}{2}=\frac{-6\pm20}{2}.$

So $c=7$ or $c=-13$.

A line of a fixed slope can be slid up and down (by changing $c$) until it just grazes the circle, and by symmetry this happens once on each side of the circle, in the direction perpendicular to the family of parallel lines. That geometric picture matches the algebra exactly: two values of $c$, one for each side.

The compact formula from the lesson was derived by substituting $y=mx+c$ into $x^2+y^2=r^2$, a circle centered at the ORIGIN.

$\Delta=4\left[r^2(1+m^2)-c^2\right].$

This circle is centered at $(1,-1)$, not the origin, so that formula cannot be used UNCHANGED here; substituting and recomputing the discriminant from scratch, as done in Parts A and B, is the method this course's toolkit supplies for a circle centered anywhere else.

Router A is centered at $(-3,0)$ and router B at $(5,0)$, so the distance between them is

$d=\sqrt{(5-(-3))^2+(0-0)^2}=\sqrt{64}=8.$

Both radii equal $5$, so $r_1+r_2=10$ and $|r_1-r_2|=0$. Since $0<8<10$, the distance lies strictly between those two bounds, so the boundaries cross at exactly two points.

Expand both equations into general form.

$(x+3)^2+y^2=25 \;\Longrightarrow\; x^2+y^2+6x-16=0,$ $(x-5)^2+y^2=25 \;\Longrightarrow\; x^2+y^2-10x=0.$

Subtract the second from the first; the $x^2+y^2$ terms cancel.

$16x-16=0 \;\Longrightarrow\; x=1.$

Substitute $x=1$ into either circle, say router B's general form.

$1+y^2-10=0 \;\Longrightarrow\; y^2=9 \;\Longrightarrow\; y=\pm3.$

The crossing points are $(1,3)$ and $(1,-3)$.

General perpendicularity, true for any two intersecting circles: subtracting one general-form equation from the other, exactly as in Part B, leaves a line $Ax+By+C=0$ with $A=D_1-D_2$ and $B=E_1-E_2$. A circle's center is $\left(-\tfrac{D}{2},-\tfrac{E}{2}\right)$, so the direction from center 1 to center 2 is

$\left(-\frac{D_2}{2}+\frac{D_1}{2},\ -\frac{E_2}{2}+\frac{E_1}{2}\right) = \left(\frac{D_1-D_2}{2},\ \frac{E_1-E_2}{2}\right),$

which is parallel to $(A,B)$, the NORMAL vector of the crossing-point line. A line's direction is always perpendicular to its own normal, so the crossing-point line is perpendicular to the line of centers for any two circles at all, not only equal-radius ones. Here $A=16$, $B=0$, matching the horizontal direction $(8,0)$ between the centers $(-3,0)$ and $(5,0)$, confirming $x=1 \perp y=0$.

Midpoint property, equal radius only: the midpoint of the two centers is $M=(1,0)$, sitting the same distance from each center. Evaluating each circle's general-form expression at $M$, the same sign-test idea used in Question 1, gives

$g_1(M)-g_2(M) = \left[\left(\tfrac{d}{2}\right)^2-r_1^2\right] - \left[\left(\tfrac{d}{2}\right)^2-r_2^2\right] = r_2^2-r_1^2.$

This difference is exactly the left side of the crossing-point line's own equation evaluated at $M$, so $M$ lies on that line precisely when $r_2^2=r_1^2$. Here $r_1=r_2=5$, so the midpoint does lie on $x=1$, but unequal radii would keep the crossing-point line perpendicular to the center line while pulling it off the midpoint.

Substitute each station's coordinates into $x^2+y^2+Dx+Ey+F=0$.

$(0,0):\ F=0, \qquad (6,0):\ 36+6D+F=0, \qquad (0,8):\ 64+8E+F=0.$

The first equation gives $F=0$ immediately. Substitute that into the other two.

$36+6D=0 \;\Longrightarrow\; D=-6, \qquad 64+8E=0 \;\Longrightarrow\; E=-8.$

Check the solved values against the two equations they came from, rather than trusting the algebra blindly.

$36+6(-6)+0 = 36-36 = 0, \qquad 64+8(-8)+0 = 64-64 = 0.$

Both equations hold, confirming $D=-6$, $E=-8$, $F=0$.

Substitute $D=-6$, $E=-8$, $F=0$ into the general form and complete the square on each variable.

$x^2-6x+y^2-8y=0 \;\Longrightarrow\; (x^2-6x+9)+(y^2-8y+16)=9+16.$

Simplify both sides.

$(x-3)^2+(y-4)^2=25.$

The center is $(3,4)$ and the radius is $\sqrt{25}=5$.

Substitute $(6,8)$ into the circle from Part B.

$(6-3)^2+(8-4)^2=9+16=25.$

This matches the right side exactly, so $(6,8)$ lies precisely on the boundary as well, and all four stations sit on the same circle.

The four stations $(0,0)$, $(6,0)$, $(6,8)$, $(0,8)$ form a rectangle: adjacent sides are horizontal and vertical, so every interior angle is a right angle. A rectangle's two diagonals are congruent and bisect each other, so both diagonals share the same midpoint, and that shared midpoint is the same distance from all four corners (half of either diagonal's length). Here the diagonal from $(0,0)$ to $(6,8)$ has midpoint

$\left(\frac{0+6}{2},\ \frac{0+8}{2}\right)=(3,4),$

which is exactly the center found in Part B, and half its length is $\sqrt{36+64}/2=5$, exactly the radius. So the shared midpoint of the diagonals IS the circle's center, at the same distance from all four corners, which is why all four stations sit on one circle. This is not a special property of these particular numbers: any rectangle's diagonals always bisect each other at a point equidistant from all four vertices, so every rectangle's vertices lie on a common circle centered there.