How do you find the center and radius of a circle from its equation?

Put it in standard form (x - h)^2 + (y - k)^2 = r^2. The center is (h, k) — the values subtracted inside the parentheses, with the signs flipped — and the radius is the square root of the number on the right side.

What does completing the square do for a circle equation?

It converts an expanded equation like x^2 + y^2 - 6x + 4y - 12 = 0 into standard form. You group the x-terms and y-terms, add the square of half each linear coefficient to both sides, and factor, revealing the center and radius.

How can you tell a circle from an ellipse on the ACT?

Look at the coefficients of x^2 and y^2. If they are equal (same sign), the graph is a circle; if they are unequal but the same sign, it is an ellipse. If only one variable is squared, it is a parabola instead.

What do a and b represent in an ellipse equation?

In x^2/a^2 + y^2/b^2 = 1, the values a and b are the semi-axes — the distances from the center to the curve along the x- and y-axes. The full axis lengths are 2a and 2b, and the longer axis lies under the larger denominator.

CramClub • Conic Sections — ACT® Guide & Practice

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

$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$

$x^2 + bx + \left(\tfrac{b}{2}\right)^2 = \left(x + \tfrac{b}{2}\right)^2$

Circles and Ellipses Live in Standard Form

On the ACT, conic sections mostly means two shapes: the circle and the ellipse. Both are defined by tidy "standard-form" equations, and nearly every question is really asking you to read the center, radius, or axis lengths straight out of that form — or to first rearrange a messy equation into it by completing the square. Master those two moves and this topic becomes almost mechanical.

The Circle

A circle is the set of all points a fixed distance $r$ (the radius) from a fixed center $(h, k)$ . Its standard equation is

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

Read it carefully: the numbers subtracted inside the parentheses are the center coordinates, and the right side is $r^2$ , not $r$ . So $(x - 3)^2 + (y + 2)^2 = 25$ has center $(3, -2)$ — note the sign flips, since $y + 2 = y - (-2)$ — and radius $\sqrt{25} = 5$ . A circle centered at the origin simplifies all the way to $x^2 + y^2 = r^2$ .

Completing the Square

The ACT often hands you a circle in expanded form, such as $x^2 + y^2 - 6x + 4y - 12 = 0$ , and asks for the center or radius. To recover standard form, complete the square in $x$ and in $y$ separately.

x^2 - 6x + y^2 + 4y &= 12 \\ (x^2 - 6x + 9) + (y^2 + 4y + 4) &= 12 + 9 + 4 \\ (x - 3)^2 + (y + 2)^2 &= 25 \end{aligned}$$ The rule for the added constant: take **half of the linear coefficient and square it**. For $-6x$, half of $-6$ is $-3$, and $(-3)^2 = 9$. Whatever you add on the left you must add on the right too. The finished form reveals center $(3, -2)$ and radius $5$. ## The Ellipse An **ellipse** is a stretched circle. Centered at the origin, its standard form is $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ Here $a$ is the distance from the center to the curve along the $x$-axis and $b$ the distance along the $y$-axis, so the full axis *lengths* are $2a$ and $2b$. The larger denominator sits under the variable along which the ellipse is longer: if $a^2 > b^2$, the ellipse is wider than it is tall. Setting $y = 0$ gives the $x$-intercepts $(\pm a, 0)$; setting $x = 0$ gives the $y$-intercepts $(0, \pm b)$. Because the denominators are $a^2$ and $b^2$, remember to take a square root before reporting a distance. ## Telling the Conics Apart You can classify a second-degree equation by its squared terms: | Both $x^2$ and $y^2$ present, and… | Shape | |---|---| | coefficients equal (same sign) | **Circle** | | coefficients unequal but same sign | **Ellipse** | | only one variable squared | **Parabola** | So $3x^2 + 3y^2 = 27$ is a circle (equal coefficients — divide by 3 to see $x^2 + y^2 = 9$), while $\tfrac{x^2}{9} + \tfrac{y^2}{4} = 1$ is an ellipse. ## A Reliable Game Plan - **Circle question?** Match to $(x - h)^2 + (y - k)^2 = r^2$; flip the inside signs for the center and square-root the right side for the radius. - **Expanded equation?** Group $x$-terms and $y$-terms, complete the square on each, and balance both sides. - **Ellipse question?** The denominators are $a^2$ and $b^2$; the semi-axes are their square roots, and the longer axis is under the bigger denominator. - **Classify?** Compare the coefficients of $x^2$ and $y^2$. ## Common Mistakes to Avoid - Reporting $r$ as the right-hand side instead of its square root (radius of $x^2 + y^2 = 16$ is $4$, not $16$). - Getting the center's signs backward: $(x + 5)^2$ means $h = -5$. - Forgetting to add the completed-square constant to **both** sides of the equation. - Confusing $a^2$ with $a$ in an ellipse — the axis length uses the square root of the denominator.

A circle has equation $(x - 4)^2 + (y + 1)^2 = 36$ . What are its center and radius?

Write the standard equation of the circle centered at $(-2, 5)$ with radius $3$ .

Find the center and radius of the circle $x^2 + y^2 - 6x + 4y - 12 = 0$ .

For the ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$ , how long are the horizontal and vertical axes?

A circle is centered at $(1, -2)$ and passes through $(4, 2)$ . Does the point $(1, 3)$ lie on the circle?

Identify the conic $x^2 + y^2 + 8x - 2y + 8 = 0$ and give its radius if it is a circle.

Reporting the radius as the right-hand side of the equation — saying the radius of $(x-1)^2 + (y-2)^2 = 49$ is $49$ .

The right side is $r^2$ . Take the square root: the radius is $\sqrt{49} = 7$ .

Getting the center signs backward, reading $(x + 3)^2 + (y - 5)^2 = 4$ as center $(3, -5)$ .

The form subtracts: $x + 3 = x - (-3)$ , so $h = -3$ ; $y - 5$ gives $k = 5$ . Center is $(-3, 5)$ .

Completing the square by adding a constant to only one side of the equation.

Whatever you add to the left (to form a perfect square) must be added to the right too, or the equation changes value.

Confusing $a^2$ with $a$ in an ellipse and reading $\tfrac{x^2}{16} + \tfrac{y^2}{9} = 1$ as a semi-axis of $16$ .

The denominator is $a^2$ . The semi-axis is $\sqrt{16} = 4$ , and the full axis is $2a = 8$ .

What is the radius of the circle given by $x^2 + y^2 = 49$ ?

What is the center of the circle $(x - 5)^2 + (y + 3)^2 = 16$ ?

The equation of a circle is $x^2 + y^2 - 10x + 6y + 18 = 0$ . What is the $x$ -coordinate of its center?

The ellipse $\dfrac{x^2}{36} + \dfrac{y^2}{16} = 1$ is graphed in the standard coordinate plane. What is the length of its longer axis?

A circle has center $(2, -1)$ and passes through the point $(6, 2)$ . What is the equation of the circle?

After completing the square, the equation $x^2 + y^2 + 4x - 8y + 4 = 0$ becomes $(x + 2)^2 + (y - 4)^2 = r^2$ . What is $r$ ?

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

$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$

$x^2 + bx + \left(\tfrac{b}{2}\right)^2 = \left(x + \tfrac{b}{2}\right)^2$

Circles and Ellipses Live in Standard Form

The Circle

A circle is the set of all points a fixed distance $r$ (the radius) from a fixed center $(h, k)$ . Its standard equation is

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

Completing the Square

x^2 - 6x + y^2 + 4y &= 12 \\ (x^2 - 6x + 9) + (y^2 + 4y + 4) &= 12 + 9 + 4 \\ (x - 3)^2 + (y + 2)^2 &= 25 \end{aligned}$$ The rule for the added constant: take **half of the linear coefficient and square it**. For $-6x$, half of $-6$ is $-3$, and $(-3)^2 = 9$. Whatever you add on the left you must add on the right too. The finished form reveals center $(3, -2)$ and radius $5$. ## The Ellipse An **ellipse** is a stretched circle. Centered at the origin, its standard form is $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ Here $a$ is the distance from the center to the curve along the $x$-axis and $b$ the distance along the $y$-axis, so the full axis *lengths* are $2a$ and $2b$. The larger denominator sits under the variable along which the ellipse is longer: if $a^2 > b^2$, the ellipse is wider than it is tall. Setting $y = 0$ gives the $x$-intercepts $(\pm a, 0)$; setting $x = 0$ gives the $y$-intercepts $(0, \pm b)$. Because the denominators are $a^2$ and $b^2$, remember to take a square root before reporting a distance. ## Telling the Conics Apart You can classify a second-degree equation by its squared terms: | Both $x^2$ and $y^2$ present, and… | Shape | |---|---| | coefficients equal (same sign) | **Circle** | | coefficients unequal but same sign | **Ellipse** | | only one variable squared | **Parabola** | So $3x^2 + 3y^2 = 27$ is a circle (equal coefficients — divide by 3 to see $x^2 + y^2 = 9$), while $\tfrac{x^2}{9} + \tfrac{y^2}{4} = 1$ is an ellipse. ## A Reliable Game Plan - **Circle question?** Match to $(x - h)^2 + (y - k)^2 = r^2$; flip the inside signs for the center and square-root the right side for the radius. - **Expanded equation?** Group $x$-terms and $y$-terms, complete the square on each, and balance both sides. - **Ellipse question?** The denominators are $a^2$ and $b^2$; the semi-axes are their square roots, and the longer axis is under the bigger denominator. - **Classify?** Compare the coefficients of $x^2$ and $y^2$. ## Common Mistakes to Avoid - Reporting $r$ as the right-hand side instead of its square root (radius of $x^2 + y^2 = 16$ is $4$, not $16$). - Getting the center's signs backward: $(x + 5)^2$ means $h = -5$. - Forgetting to add the completed-square constant to **both** sides of the equation. - Confusing $a^2$ with $a$ in an ellipse — the axis length uses the square root of the denominator.