Circles Practice Questions

The circumference of a circle is given by:

\(C = 2\pi r\)

Substitute \(r = 7\):

\(C = 2\pi(7) = 14\pi \approx 14(3.14) = 43.96 \text{ cm}\)

Choice A uses the formula \(\pi r\) (forgetting the factor of 2). Choice C gives the area of the circle (\(\pi r^2\)) instead of the circumference.

The area formula uses the radius, not the diameter. First, find the radius by dividing the diameter by 2:

\(r = \dfrac{10}{2} = 5 \text{ inches}\)

Now apply the area formula:

\(A = \pi r^2 = \pi(5)^2 = 25\pi \text{ in}^2\)

Choice B uses the diameter (10) as the radius in the formula, giving \(\pi(10)^2 = 100\pi\). Always convert diameter to radius before using the area formula.

Start with the area formula and solve for the radius:

\(\pi r^2 = 81\pi\)

Divide both sides by \(\pi\):

\(r^2 = 81\)

Take the square root of both sides:

\(r = 9 \text{ cm}\)

Choice B gives the diameter, not the radius. Always check whether a question is asking for radius or diameter.

The arc length formula relates the central angle to the full circumference:

\(s = \dfrac{\theta}{360°} \cdot 2\pi r\)

Substitute \(\theta = 60°\) and \(r = 12\):

\(s = \dfrac{60°}{360°} \cdot 2\pi(12) = \dfrac{1}{6} \cdot 24\pi = 4\pi \text{ cm}\)

The central angle is \(\frac{1}{6}\) of a full 360° rotation, so the arc is \(\frac{1}{6}\) of the full circumference.

The area of a sector is a fraction of the full circle’s area, based on the central angle:

\(A_{\text{sector}} = \dfrac{\theta}{360°} \cdot \pi r^2\)

Substitute \(\theta = 90°\) and \(r = 6\):

\(A = \dfrac{90°}{360°} \cdot \pi(6)^2 = \dfrac{1}{4} \cdot 36\pi = 9\pi \text{ cm}^2\)

A 90° sector is exactly \(\frac{1}{4}\) of the circle (a quarter circle). Choice D gives the area of the entire circle.

The standard form of a circle’s equation is:

\((x – h)^2 + (y – k)^2 = r^2\)

where \((h, k)\) is the center and \(r\) is the radius. Substitute \(h = 2\), \(k = -3\), and \(r = 5\):

\((x – 2)^2 + (y – (-3))^2 = 5^2\)

\((x – 2)^2 + (y + 3)^2 = 25\)

Be careful with signs: subtracting \(-3\) becomes \(+3\). Also, the right side is \(r^2\), not \(r\), so 5 becomes 25 (choice C makes this mistake).

Compare the given equation to the standard form \((x – h)^2 + (y – k)^2 = r^2\):

• \((x + 4)^2 = (x – (-4))^2\), so \(h = -4\)
• \((y – 1)^2\), so \(k = 1\)
• \(r^2 = 49\), so \(r = 7\)

The center is \((-4, 1)\) and the radius is 7. Choice B flips both signs of the center. Choice C fails to take the square root, giving \(r = 49\) instead of 7.

The inscribed angle theorem states that an inscribed angle is half the measure of its intercepted arc:

\(\text{inscribed angle} = \dfrac{1}{2} \cdot \text{arc}\)

Substitute 80° for the arc:

\(\dfrac{1}{2}(80°) = 40°\)

Remember: An inscribed angle has its vertex on the circle, while a central angle has its vertex at the center. A central angle equals its intercepted arc, but an inscribed angle equals half its intercepted arc.

This is known as Thales’ Theorem: An inscribed angle that intercepts a diameter is always a right angle (90°), regardless of where the third point is located on the circle.

This follows from the inscribed angle theorem. A diameter intercepts an arc of 180° (a semicircle), so the inscribed angle is:

\(\dfrac{1}{2}(180°) = 90°\)

This property is useful in many geometry problems. Any triangle inscribed in a semicircle with the diameter as one side is automatically a right triangle.

A fundamental property of circles: a tangent line is always perpendicular to the radius drawn to the point of tangency. So the angle between them is 90°.

This property is useful in many problems involving tangent lines. For example, if a triangle is formed by a tangent segment, a radius, and a line from the external point to the center, it is always a right triangle with the right angle at the point of tangency.

Complete the square for both \(x\) and \(y\) to convert to standard form. Start by grouping the \(x\) and \(y\) terms:

\((x^2 – 6x) + (y^2 + 4y) = 12\)

For \(x^2 – 6x\), take half of −6 (which is −3), square it (9), and add:

\((x^2 – 6x + 9) = (x – 3)^2\)

For \(y^2 + 4y\), take half of 4 (which is 2), square it (4), and add:

\((y^2 + 4y + 4) = (y + 2)^2\)

Add the same values to the right side to keep the equation balanced:

\((x – 3)^2 + (y + 2)^2 = 12 + 9 + 4 = 25\)

The center is \((3, -2)\) and the radius is \(\sqrt{25} = 5\).

When a circle is inscribed in a square, the diameter of the circle equals the side length of the square. So:

\(d = 8 \text{ cm} \implies r = 4 \text{ cm}\)

Apply the area formula:

\(A = \pi r^2 = \pi(4)^2 = 16\pi \text{ cm}^2\)

Choice C uses the side length (8) as the radius instead of the diameter.

First, find the radius using the circumference formula:

\(C = 2\pi r \implies 18\pi = 2\pi r \implies r = 9 \text{ cm}\)

Now use the radius to find the area:

\(A = \pi r^2 = \pi(9)^2 = 81\pi \text{ cm}^2\)

Choice B would result from using \(r = 18\) directly instead of solving for \(r\) first.

When two secants meet at a point outside the circle, the angle formed equals half the difference of the intercepted arcs:

\(\text{angle} = \dfrac{1}{2}|\text{far arc} – \text{near arc}|\)

Substitute the two arcs:

\(\dfrac{1}{2}(120° – 40°) = \dfrac{1}{2}(80°) = 40°\)

Contrast this with angles formed inside the circle (intersecting chords), which equal half the sum of the intercepted arcs.

Calculate the distance from the point \((3, 4)\) to the center \((0, 0)\) using the distance formula:

\(d = \sqrt{(3 – 0)^2 + (4 – 0)^2} = \sqrt{9 + 16} = \sqrt{25} = 5\)

Since the distance equals the radius, the point lies on the circle. In general:

• If \(d \lt r\), the point is inside the circle.
• If \(d = r\), the point is on the circle.
• If \(d \gt r\), the point is outside the circle.

You can also verify by plugging \((3, 4)\) into the circle’s equation:

\(x^2 + y^2 = 25 \implies 9 + 16 = 25\)