Circles: Additional Practice (Set 2)

Substitute the line into the circle, replacing $y$ by $2x$ everywhere.

$x^2 + (2x)^2 = 20 \;\Longrightarrow\; 5x^2 = 20 \;\Longrightarrow\; x = \pm 2$

Two real roots means two intersection points. Feed each root back into the line to recover its $y$: $x = 2$ gives $y = 4$, and $x = -2$ gives $y = -4$.

The points are $(2, 4)$ and $(-2, -4)$. The pair $(4, 2)$ and $(-4, -2)$ does lie on the circle, but not on the line $y = 2x$, so it fails half the requirement.

A shared point satisfies both equations, so it satisfies their difference. Subtracting cancels the $x^2$ and $y^2$ terms and leaves a linear equation.

$(-4x - 2y - 11) - (-2x + 2y - 3) = -2x - 4y - 8 = 0$

Divide by $-2$ to simplify.

$x + 2y + 4 = 0$

Check it: the point $(2, -3)$ satisfies both circles ($4 + 9 - 8 + 6 - 11 = 0$ and $4 + 9 - 4 - 6 - 3 = 0$), and it satisfies this line too, since $2 - 6 + 4 = 0$.

Tangency to a vertical line means the circle touches it exactly once, so the radius equals the horizontal distance from the center to that line.

$r = |7 - 2| = 5$

Substitute the center $(2, 3)$ and $r^2 = 25$ into the center-radius form.

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

A quick check: substituting $x = 7$ gives $(y - 3)^2 = 0$, a double root at $y = 3$, so the line touches at the single point $(7, 3)$.

The squared terms are negative, so multiply the whole equation by $-1$ to make their coefficients $+1$.

$x^2 + y^2 - 4x - 6y + 3 = 0$

Complete both squares. Half of $-4$ is $-2$ (square $4$), and half of $-6$ is $-3$ (square $9$).

$(x - 2)^2 + (y - 3)^2 = -3 + 4 + 9 = 10$

The right side is $r^2 = 10$, so the radius is $\sqrt{10}$, which does not simplify further. The center is $(2, 3)$.

The center is the midpoint of the diameter, so average the coordinates: $\left(\frac{-3 + 7}{2}, \frac{5 + (-1)}{2}\right) = (2, 2)$.

The radius is half the distance between the endpoints. The gaps are $10$ across and $6$ down, so the squared diameter is $100 + 36 = 136$, and $r^2$ is a quarter of that (halving a length quarters its square).

$r^2 = \frac{136}{4} = 34 \;\Longrightarrow\; (x - 2)^2 + (y - 2)^2 = 34$

Expand to reach the general form, where the constant is $4 + 4 - 34 = -26$.

$x^2 + y^2 - 4x - 4y - 26 = 0$

Passing through the origin means $(0, 0)$ satisfies the equation, so substitute it and see what is left.

$0 + 0 - 0 + 0 + F = 0 \;\Longrightarrow\; F = 0$

Now confirm the graph really is a circle and not a degenerate case, using the sign of $D^2 + E^2 - 4F$ with $D = -10$ and $E = 6$.

$100 + 36 - 0 = 136 > 0$

So $F = 0$ gives an honest circle, centered at $(5, -3)$ with radius $\frac{\sqrt{136}}{2} = \sqrt{34}$, and it passes through the origin. The choice $F = 34$ makes the test value zero, which collapses the graph to the single point $(5, -3)$.

Substitute $x = a$ and see when the resulting equation has no real solution for $y$.

$(y + 2)^2 = 9 - (a - 1)^2$

A square of a real number is never negative, so there is no solution exactly when the right side is negative, that is when $(a - 1)^2 > 9$, or $|a - 1| > 3$.

$a - 1 < -3 \;\text{ or }\; a - 1 > 3 \;\Longrightarrow\; a < -2 \;\text{ or }\; a > 4$

Geometrically the circle is centered at $x = 1$ with radius $3$, so it occupies the strip from $x = -2$ to $x = 4$. At the two endpoints the line is tangent, so those values must be excluded from a miss.

All four lines pass through $(5, 7)$, so passing through the point is not enough: the tangent is the one that meets the circle only there, which means the substituted quadratic must have a double root.

Test $3x + 4y = 43$ by writing $x = \frac{43 - 4y}{3}$ and substituting, then clearing the denominator.

$(37 - 4y)^2 + 9(y - 3)^2 = 225 \;\Longrightarrow\; 25y^2 - 350y + 1225 = 0 \;\Longrightarrow\; (y - 7)^2 = 0$

The repeated root $y = 7$ confirms a single touch at $(5, 7)$. The radius from $(2, 3)$ to $(5, 7)$ has slope $\frac{4}{3}$, and the tangent line's slope is $-\frac{3}{4}$, the perpendicular value, exactly as expected. The line $4x - 3y = -1$ has slope $\frac{4}{3}$, so it runs along the radius and cuts straight through the circle.

Substitute each point into $x^2 + y^2 + Dx + Ey + F = 0$, which is linear in the three unknowns.

$D + F = -1, \qquad E + F = -1, \qquad 3D + 2E + F = -13$

The first two force $D = E$. Substituting $E = D$ and $F = -1 - D$ into the third gives $3D + 2D - 1 - D = -13$, so $4D = -12$ and $D = -3$. Then $E = -3$ and $F = 2$.

$x^2 + y^2 - 3x - 3y + 2 = 0 \;\Longrightarrow\; \text{center } \left(-\frac{D}{2}, -\frac{E}{2}\right) = \left(\frac{3}{2}, \frac{3}{2}\right)$

The average of the three points is $\left(\frac{4}{3}, 1\right)$, which is a tempting answer and a wrong one: the center of a circle is not the average of points on it.

Substitute $x = 0$ and $y = 0$ into the equation and see what survives.

$0^2 + 0^2 + D(0) + E(0) + F = F$

So the origin satisfies the equation exactly when $F = 0$, whatever $D$ and $E$ may be. That statement is always true.

The others fail. Completing the square puts the center at $\left(-\frac{D}{2}, -\frac{E}{2}\right)$, not $(D, E)$; the radius is $\frac{1}{2}\sqrt{D^2 + E^2 - 4F}$, not $\sqrt{F}$; and when $D^2 + E^2 - 4F$ is zero or negative the graph is a single point or nothing at all, so it is not always a circle.

Write the condition with the distance formula, then square both sides to clear the roots.

$\sqrt{x^2 + y^2} = 2\sqrt{(x - 3)^2 + y^2} \;\Longrightarrow\; x^2 + y^2 = 4\left[(x - 3)^2 + y^2\right]$

Expand the right side and gather everything on one side: $x^2 + y^2 = 4x^2 - 24x + 36 + 4y^2$, so $3x^2 + 3y^2 - 24x + 36 = 0$. Divide by $3$ and complete the square.

$x^2 + y^2 - 8x + 12 = 0 \;\Longrightarrow\; (x - 4)^2 + y^2 = 4$

The center is $(4, 0)$ and the radius is $2$. Check the point $(6, 0)$: it is $6$ from the origin and $3$ from $(3, 0)$, which matches the rule, and it does lie on this circle.

Concentric means same center, so find the center of the given circle with $D = -4$ and $E = 10$.

$\left(-\frac{-4}{2},\; -\frac{10}{2}\right) = (2, -5)$

Now write the center-radius form with that center and $r = 7$, so the right side is $r^2 = 49$.

$(x - 2)^2 + (y + 5)^2 = 49$

The radius of the original circle ($3$, from $16 + 100 - 80 = 36$) plays no part: only the center is copied.