Circles: worksheet with answers

A circle written in standard form has the equation \((x-h)^{2}+(y-k)^{2}=r^{2}\), where \(r\) is the radius. In the equation \((x+4)^{2}+(y-9)^{2}=49\), the value of \(r^{2}\) is 49. Therefore, \(r=\sqrt{49}=7\). The values 4 and 9 come from the center coordinates, and 49 is the value of \(r^{2}\), not the radius itself.

A circle in standard form is written as \((x-h)^{2}+(y-k)^{2}=r^{2}\), where \(r\) is the radius. In the equation \((x+4)^{2}+(y-9)^{2}=36\), the value of \(r^{2}\) is 36. Therefore, \(r=\sqrt{36}=6\).

Choice A uses the x-coordinate of the center, choice B uses the y-coordinate of the center, and choice D incorrectly uses \(r^{2}\) instead of the radius.

The sine function has period \(2\pi\). Since \(2\pi=\frac{12\pi}{6}\), reduce the angle by subtracting multiples of \(\frac{12\pi}{6}\):

\(\frac{53\pi}{6}-\frac{48\pi}{6}=\frac{5\pi}{6}\).

Therefore, \(\sin\left(\frac{53\pi}{6}\right)=\sin\left(\frac{5\pi}{6}\right)\). The angle \(\frac{5\pi}{6}\) is in Quadrant II, where sine is positive, and its reference angle is \(\frac{\pi}{6}\). Since \(\sin\left(\frac{\pi}{6}\right)=\frac{1}{2}\), the exact value is \(\frac{1}{2}\).

A circle in standard form is written as \((x-h)^{2}+(y-k)^{2}=r^{2}\), where \(r\) is the radius. In the equation \((x+4)^{2}+(y-9)^{2}=49\), the value of \(r^{2}\) is 49. Therefore, \(r=\sqrt{49}=7\).

Choice A uses the x-coordinate of the center, choice B uses the y-coordinate of the center, and choice D incorrectly uses \(r^{2}\) instead of the radius.

A circle in standard form is written as \((x-h)^{2}+(y-k)^{2}=r^{2}\), where the center is \((h,k)\). In the equation \((x+6)^{2}+(y-4)^{2}=25\), the expression \(x+6\) can be rewritten as \(x-(-6)\), so \(h=-6\). The expression \(y-4\) shows that \(k=4\). Therefore, the center is \((-6,4)\).

A circle in standard form is written as \((x-h)^{2}+(y-k)^{2}=r^{2}\), where the center is \((h,k)\). In the equation \((x+6)^{2}+(y-4)^{2}=25\), the expression \(x+6\) can be rewritten as \(x-(-6)\), so \(h=-6\). The expression \(y-4\) shows that \(k=4\). Therefore, the center is \((-6,4)\).

To convert radians to degrees, multiply by \(\frac{180}{\pi}\).

\(\frac{13\pi}{18}\times\frac{180}{\pi}=\frac{13\times180}{18}\)

The \(\pi\) terms cancel, and \(180\div18=10\).

\(13\times10=130\), so the angle measures \(130^\circ\).

The minor arc RS and the major arc RTS have the same endpoints, so together they make the entire circle. Therefore, the circumference is the sum of the two arc lengths:

\(47+186=233\)

So, the circumference of the circle is 233 units.

The original circle has center \((-4,1)\) because the equation is in the form \((x-h)^{2}+(y-k)^{2}=r^{2}\). Translating the circle 5 units to the right adds 5 to the x-coordinate of the center: \(-4+5=1\). Translating 3 units down subtracts 3 from the y-coordinate: \(1-3=-2\). The new center is \((1,-2)\). The radius does not change, so \(r^{2}=49\) remains the same. Using standard form, the translated circle is \((x-1)^{2}+(y+2)^{2}=49\).

A circle in standard form \((x-h)^{2}+(y-k)^{2}=r^{2}\) has center \((h,k)\) and radius \(r\). A circle intersects the y-axis at exactly one point when it is tangent to the y-axis, so the horizontal distance from the center to the y-axis must equal the radius. Since the y-axis is \(x=0\), this condition is \(|h|=r\).

Check each choice:

• A: Center \((4,-1)\), radius \(3\). Since \(|4|\ne 3\), not tangent.
• B: Center \((-2,5)\), radius \(5\). Since \(|-2|\ne 5\), not tangent.
• C: Center \((-6,2)\), radius \(6\). Since \(|-6|=6\), the circle is tangent to the y-axis and intersects it at exactly one point.
• D: Center \((1,-4)\), radius \(4\). Since \(|1|\ne 4\), not tangent to the y-axis.

Therefore, choice C is correct.

In the standard form of a circle equation, the number on the right side is the radius squared. For Circle A, \((x+4)^{2}+(y-7)^{2}=13\), so the radius squared is 13.

Translations change the location of a circle but do not change its radius. Therefore, Circle B has the same radius squared, so \(a=13\).

Now evaluate the expression:

\(4a-3=4(13)-3=52-3=49\).

In the standard form of a circle, \((x-p)^{2}+(y-q)^{2}=r^{2}\), the number on the right side is the radius squared.

For Circle A, the equation is \((x-4)^{2}+(y+1)^{2}=13\), so the radius squared is 13.

A translation changes the location of a circle but does not change its radius. Therefore, Circle B also has radius squared equal to 13, which means \(a=13\).

Now evaluate the expression:

\(4a-3=4(13)-3=52-3=49\)