Aimathic – Equation of a circle – Free Math Worksheets

Equation of a circle

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Difficulty: 1 – 5

53357010

The unit circle has equation \(x^2+y^2=1\). Points \(P(0.8, 0.6)\) and \(Q(1, 1)\) are shown. Determine algebraically which point lies on the circle.

Hints

- A point lies on a curve when its coordinates satisfy the curve’s equation. - Square the x-coordinate and y-coordinate, add them, and compare the sum with \(1\).

Solution

1. Substitute \(P(0.8, 0.6)\): \((0.8)^2+(0.6)^2=0.64+0.36=1\). Therefore, \(P\) lies on the circle. 2. Substitute \(Q(1, 1)\): \(1^2+1^2=2\ne 1\). Therefore, \(Q\) does not lie on the circle.

Answer

Only \(P(0.8, 0.6)\) lies on the unit circle.

53490710

A circular island is modeled on a coordinate plane by a circle centered at \((0, 0)\) with radius \(10\,\text{km}\). A straight power line follows the line \(y=6\), shown as a dashed line. Find the coordinates of the two points where the power line crosses the island's shoreline.

Hints

- Each point must lie on both the circle and the dashed line. - Substitute the line's fixed \(y\)-coordinate into the equation of the circle.

Solution

1. The circle has equation \(x^2+y^2=100\). 2. At each intersection, \(y=6\), so \(x^2+6^2=100\). 3. Thus, \(x^2=64\), giving \(x=8\) or \(x=-8\). 4. The intersection points are \((8, 6)\) and \((-8, 6)\).

Answer

\((8, 6)\) and \((-8, 6)\)

52864710

A point \(P(x, y)\) lies on the unit circle in Quadrant I, with coordinates \((\cos(\alpha), \sin(\alpha))\). 1. Find the missing x-coordinate when \(y=0.44\). Round to the nearest thousandth. 2. Find the exact y-coordinate when \(x=\frac{3}{5}\). 3. Determine whether \(Q(0.6, 0.7)\) can lie on the unit circle. Justify your answer.

Hints

- Every point on the unit circle is a distance \(1\) from the origin. - Use the Pythagorean theorem to relate \(x\) and \(y\). - A point lies on the circle only if its coordinates satisfy \(x^2+y^2=1\).

Solution

1. Use \(x^2+y^2=1\): \(x^2+0.44^2=1\), so \(x^2=0.8064\). Because \(P\) is in Quadrant I, \(x=\sqrt{0.8064}\approx0.898\). 2. Substitute \(x=\frac{3}{5}\): \(\frac{9}{25}+y^2=1\), so \(y^2=\frac{16}{25}\). In Quadrant I, \(y=\frac{4}{5}\). 3. For \(Q\), \(0.6^2+0.7^2=0.36+0.49=0.85\ne1\). Therefore, \(Q\) is not on the unit circle.

Answer

1. \(x\approx0.898\) 2. \(y=\frac{4}{5}\) 3. No, because \(0.6^2+0.7^2=0.85\ne1\).

52879310

Consider all points \((x, y)\) that satisfy \((x-3)^2+(y+1)^2=16\). a) Interpret the equation geometrically. State the center and radius. b) Solve the equation for \(y\). Use your result to explain why the entire figure is not the graph of a function that assigns one y-value to each x-value. c) Give two subsets of the figure that can each be viewed as the graph of a function. State the domain of one of these functions.

Hints

- Compare the equation with the standard form of a circle. - A graph represents a function of \(x\) only when each x-value has at most one y-value. - Solving a squared equation introduces both a positive and a negative square root. - Require the expression under the square root to be nonnegative.

Solution

1. Compare the equation with \((x-h)^2+(y-k)^2=r^2\). The figure is a circle with center \((3, -1)\) and radius \(4\). 2. Solve for \(y\): \((y+1)^2=16-(x-3)^2\), so \(y=-1\pm\sqrt{16-(x-3)^2}\). 3. For most x-values between \(-1\) and \(7\), the plus and minus signs give two different y-values. Therefore, the entire circle fails the vertical line test and is not the graph of a function of \(x\). 4. The upper semicircle is \(f(x)=-1+\sqrt{16-(x-3)^2}\), and the lower semicircle is \(g(x)=-1-\sqrt{16-(x-3)^2}\). 5. For either function, the radicand must be nonnegative: \(16-(x-3)^2\ge 0\). Thus, \((x-3)^2\le 16\), giving the domain \([-1, 7]\).

Answer

a) A circle with center \((3, -1)\) and radius \(4\) b) \(y=-1\pm\sqrt{16-(x-3)^2}\). The circle is not a function of \(x\) because many x-values correspond to two y-values. c) The upper and lower semicircles are \(f(x)=-1+\sqrt{16-(x-3)^2}\) and \(g(x)=-1-\sqrt{16-(x-3)^2}\). Each has domain \([-1, 7]\).

52879410

Consider \(g(x)=-\sqrt{64-x^2}\). a) Find the maximal domain and the range of \(g\). b) Describe the geometric shape of the graph. c) Determine whether the graph is symmetric about the y-axis. d) Show that every point on the graph satisfies \(x^2+y^2=64\). Explain why not every point satisfying that equation lies on the graph of \(g\).

Hints

- Require the expression under the square root to be nonnegative. - Account for the negative sign in front of the square root when finding the range. - Test y-axis symmetry by comparing \(g(-x)\) and \(g(x)\). - Squaring can remove sign information.

Solution

1. The radicand must be nonnegative: \(64-x^2\ge 0\). Thus, \(x^2\le 64\), so the domain is \([-8, 8]\). 2. Because \(\sqrt{64-x^2}\ge 0\), the negative sign makes \(g(x)\le 0\). The minimum value is \(g(0)=-8\), and the maximum value is \(0\) at \(x=\pm8\). Therefore, the range is \([-8, 0]\). 3. The graph is the lower semicircle centered at the origin with radius \(8\). 4. Since \(g(-x)=-\sqrt{64-(-x)^2}=g(x)\), the function is even and its graph is symmetric about the y-axis. 5. If \(y=-\sqrt{64-x^2}\), then squaring gives \(y^2=64-x^2\), so \(x^2+y^2=64\). The circle equation also includes points with \(y>0\), but \(g(x)\le 0\), so those points are not on the graph of \(g\).

Answer

a) Domain: \([-8, 8]\); range: \([-8, 0]\) b) The lower semicircle centered at the origin with radius \(8\) c) Yes. \(g(-x)=g(x)\), so the graph is symmetric about the y-axis. d) Squaring gives \(x^2+y^2=64\). The converse fails for points on the upper semicircle, where \(y>0\).

53490610

A circle centered at \(M(0, 0)\) has radius \(5\) units. 1. Find the points where the circle intersects the x-axis. 2. Find the y-coordinates of the points on the circle whose x-coordinate is \(4\).

Hints

- Use the equation of a circle centered at the origin. - Points on the x-axis have y-coordinate \(0\). - A vertical line can intersect a circle at two points.

Solution

1. The circle has equation \(x^2 + y^2 = 25\). 2. On the x-axis, \(y = 0\), so \(x^2 = 25\). Thus, the intercepts are \((5, 0)\) and \((-5, 0)\). 3. For \(x = 4\), substitute into the equation: \(4^2 + y^2 = 25\), so \(y^2 = 9\). 4. Therefore, \(y = 3\) or \(y = -3\).

Answer

1. \((5, 0)\) and \((-5, 0)\) 2. \(y = 3\) and \(y = -3\)

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