Circles: Practice

Match the equation against the center-radius form $(x - h)^2 + (y - k)^2 = r^2$. The parentheses are $x - 3$ and $y - 5$, so $h = 3$ and $k = 5$.

The right side is $r^2$, not $r$, so take its square root.

$r = \sqrt{49} = 7$

The center is $(3, 5)$ and the radius is $7$.

The form subtracts the center coordinates, so rewrite the first parenthesis as a subtraction: $x + 2 = x - (-2)$, which gives $h = -2$. The second is $y - 6$, so $k = 6$.

The radius is the square root of the right side.

$r = \sqrt{9} = 3$

The center is $(-2, 6)$ and the radius is $3$. A plus sign inside the parentheses always signals a negative coordinate.

With center $(0, 0)$ the subtractions disappear, and the center-radius form becomes $x^2 + y^2 = r^2$. The right side is the radius squared.

$x^2 + y^2 = 6^2 = 36$

So the equation is $x^2 + y^2 = 36$. The choice $x^2 + y^2 + 36 = 0$ says $x^2 + y^2 = -36$, which no real point satisfies.

Substitute $h = 4$, $k = -1$, and $r = 2$ into $(x - h)^2 + (y - k)^2 = r^2$, watching the double negative in the $y$-term: $y - (-1) = y + 1$.

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

The right side must be $r^2 = 4$, not the radius $2$.

A bare $x^2$ means $(x - 0)^2$, so $h = 0$. The second parenthesis is $y - 4$, so $k = 4$, which puts the center on the $y$-axis.

$r = \sqrt{16} = 4$

The center is $(0, 4)$ and the radius is $4$.

Group the $x$-terms and the $y$-terms and move the constant to the right: $(x^2 - 4x) + (y^2 - 6y) = -9$.

Half of $-4$ is $-2$, whose square is $4$; half of $-6$ is $-3$, whose square is $9$. Add both constants to both sides.

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

The center is $(2, 3)$ and the radius is $\sqrt{4} = 2$.

The radius is the distance from the center to any point on the circle. The two points share an $x$-coordinate, so the segment joining them is vertical and its length is the gap between the $y$-coordinates.

$r = \sqrt{(2 - 2)^2 + (1 - (-3))^2} = \sqrt{0 + 16} = 4$

The radius is $4$. The value $16$ is $r^2$, and $8$ would be the diameter.

On the $x$-axis every point has $y = 0$, so substitute $y = 0$ and solve for $x$.

$x^2 + 0^2 = 49 \;\Longrightarrow\; x = \pm 7$

The positive crossing is $(7, 0)$. That makes sense geometrically: the circle is centered at the origin with radius $\sqrt{49} = 7$, so it meets each axis exactly $7$ units from the center.

The left side is a sum of two squares of real numbers, and each square is greater than or equal to zero. A sum like that can equal zero only if both squares are zero at once.

$(x - 4)^2 = 0 \;\text{ and }\; (y + 1)^2 = 0 \;\Longrightarrow\; x = 4,\; y = -1$

Exactly one point satisfies the equation, namely $(4, -1)$. This is the degenerate case: the circle has shrunk to its own center.

For real numbers $x$ and $y$, both $x^2$ and $y^2$ are greater than or equal to zero, so their sum is never negative.

$x^2 + y^2 \geq 0 > -16$

No real point can satisfy the equation, so the graph is empty. Writing $r = \sqrt{-16}$ is not a radius; it is a signal that nothing gets drawn.

Group and complete both squares. Half of $-2$ is $-1$ (square $1$), and half of $8$ is $4$ (square $16$), so add $1$ and $16$ to both sides of $(x^2 - 2x) + (y^2 + 8y) = -8$.

$(x - 1)^2 + (y + 4)^2 = -8 + 1 + 16 = 9$

The right side is $r^2 = 9$, so the radius is $\sqrt{9} = 3$. The center, for the record, is $(1, -4)$.

Substitute $h = 0$, $k = -3$, and $r = \sqrt{5}$ into $(x - h)^2 + (y - k)^2 = r^2$. The $y$-term becomes $y - (-3) = y + 3$, and the right side is the radius squared.

$x^2 + (y + 3)^2 = \left(\sqrt{5}\right)^2 = 5$

Squaring the radical removes it, so the right side is the plain number $5$.