Free math worksheets

- A point lies on a curve when its coordinates satisfy the curve's equation. - Test the two points separately in \(x^2+y^2=1\). - Compare each resulting value with the right side of the equation.

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 \(Q(1, 1)\): \(1^2+1^2=2\ne1\). Therefore, \(Q\) does not lie on the circle.

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

- Match each squared binomial to \(x-h\) or \(y-k\). - Pay attention to the sign inside the y-binomial. - The number on the right is \(r^2\), not \(r\).

1. Compare the equation with standard form \((x-h)^2+(y-k)^2=r^2\). 2. Since \(x-4=x-h\), \(h=4\). Since \(y+3=y-(-3)\), \(k=-3\). 3. Since \(r^2=49\), the radius is \(r=7\).

Center: \((4,-3)\); radius: \(7\)

- How does the center-radius form simplify when the center is the origin? - What y-coordinate must an x-axis intersection have? - For part c), substitute the specified x-coordinate into the equation you wrote and keep both square roots.

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 \(x=2\), substitute into the equation: \(2^2+y^2=25\), so \(y^2=21\). 4. Therefore, \(y=\sqrt{21}\) or \(y=-\sqrt{21}\).

a) \(x^2+y^2=25\) b) \((5,0)\) and \((-5,0)\) c) \(y=\sqrt{21}\) and \(y=-\sqrt{21}\)

- Use the center and radius to write the circle in center-radius form. - At an intersection, the point must have the power line's fixed y-coordinate and also satisfy the circle equation. - Keep both square roots when solving for \(x\).

1. The circle centered at the origin with radius \(10\) has equation \(x^2+y^2=100\). 2. At an intersection with the power line, \(y=5\). Substitute: \(x^2+5^2=100\). 3. Thus \(x^2=75\), so \(x=5\sqrt{3}\) or \(x=-5\sqrt{3}\). 4. The intersection points are \((5\sqrt{3},5)\) and \((-5\sqrt{3},5)\).

a) \(x^2+y^2=100\) b) \((5\sqrt{3},5)\) and \((-5\sqrt{3},5)\)

- Use the wheel's center and radius in the standard equation of a circle. - Substitute the specified height for \(y\) before solving for the horizontal coordinate. - The two square roots represent positions on opposite sides of the vertical centerline.

1. The circle has center \((0,20)\) and radius \(18\), so its equation is \(x^2+(y-20)^2=324\). 2. At a height of \(29\,\text{ft}\), substitute \(y=29\): \(x^2+9^2=324\). 3. Thus, \(x^2=243\), so \(x=\pm9\sqrt{3}\). 4. Therefore, the rider is about \(15.6\,\text{ft}\) to either side of the centerline.

a) \(x^2+(y-20)^2=324\) b) \(x=\pm9\sqrt{3}\,\text{ft}\), approximately \(15.6\,\text{ft}\) to either side of the centerline

- Use the leftmost and rightmost points of the circle to locate the horizontal midpoint. - Check the highest and lowest points to locate the vertical midpoint and radius. - Once you know \((h,k)\) and \(r\), use the standard center-radius form of a circle.

1. The horizontal extremes are at \(x=-5\) and \(x=1\), so their midpoint has x-coordinate \(-2\). The vertical extremes are at \(y=-2\) and \(y=4\), so their midpoint has y-coordinate \(1\). 2. Therefore, the center is \((-2,1)\). 3. The distance from the center to any extreme point is \(3\), so the radius is \(3\). 4. Substitute \(h=-2\), \(k=1\), and \(r=3\) into \((x-h)^2+(y-k)^2=r^2\): \((x+2)^2+(y-1)^2=9\).

a) Center: \((-2,1)\); radius: \(3\) b) \((x+2)^2+(y-1)^2=9\)

- Rewrite each binomial mentally in the form \(x-h\) or \(y-k\). - Ask what value of \(h\) makes \(x-h=x+5\). - In standard form, what does the constant on the right represent: \(r\) or \(r^2\)?

1. Standard form is \((x-h)^2+(y-k)^2=r^2\). 2. The expression \(x+5\) is \(x-(-5)\), so the center's x-coordinate is \(-5\), not \(5\). The expression \(y-2\) gives y-coordinate \(2\), not \(-2\). 3. The right side is \(r^2=16\), so \(r=4\), not \(16\). 4. The correct center is \((-5,2)\) and the radius is \(4\).

The signs inside the binomials are opposite the center coordinates, and the right side is the radius squared. The correct center is \((-5,2)\) and the radius is \(4\).

- Apply the zero-product property to the second equation. - Substitute each resulting fixed value into the other equation. - List each distinct ordered pair only once.

1. By the zero-product property, \(x-3=0\) or \(y-4=0\). Thus, consider the cases \(x=3\) and \(y=4\). 2. If \(x=3\), then \(9+y^2=25\), so \(y=4\) or \(y=-4\). This gives \((3,4)\) and \((3,-4)\). 3. If \(y=4\), then \(x^2+16=25\), so \(x=3\) or \(x=-3\). This gives \((3,4)\) and \((-3,4)\). 4. Remove the repeated ordered pair \((3,4)\).

\(\{(3,4),(3,-4),(-3,4)\}\)

- Solve the linear equation for one variable. - Substitute into the quadratic equation and expand. - Factor the resulting quadratic, then find each matching coordinate.

1. Solve the linear equation for \(x\): \(x=y+5\). 2. Substitute into the quadratic equation: \((y+5)^2+y^2=13\). 3. Expand and simplify: \(2y^2+10y+12=0\). Divide by \(2\): \(y^2+5y+6=0\). 4. Factor: \((y+2)(y+3)=0\). Thus, \(y=-2\) or \(y=-3\). 5. Using \(x=y+5\), the corresponding values are \(x=3\) and \(x=2\).

\(\{(3,-2),(2,-3)\}\)

- State the value excluded by the denominator. - Solve the rational equation for one variable. - Substitute into the quadratic equation, then verify the domain restriction.

1. The denominator requires \(y\ne1\). 2. Solve the first equation for \(x\): \(x+2=2(y-1)\), so \(x=2y-4\). 3. Substitute into the second equation: \((2y-4)^2+y^2=13\). 4. Expand and simplify: \(5y^2-16y+3=0\). 5. Factor: \((5y-1)(y-3)=0\). Thus, \(y=\frac{1}{5}\) or \(y=3\). 6. Using \(x=2y-4\), the corresponding values are \(x=-\frac{18}{5}\) and \(x=2\). Both y-values satisfy \(y\ne1\).

\(\left\{\left(-\frac{18}{5},\frac{1}{5}\right),(2,3)\right\}\)

- Solve the linear equation for one variable. - Substitute into the second equation. - Factor the resulting quadratic and find each corresponding coordinate.

1. Solve the linear equation for \(y\): \(y=1-x\). 2. Substitute into the second equation: \(x^2+(1-x)^2=13\). 3. Expand and simplify: \(2x^2-2x-12=0\). Divide by \(2\): \(x^2-x-6=0\). 4. Factor: \((x-3)(x+2)=0\). Thus, \(x=3\) or \(x=-2\). 5. Using \(y=1-x\), the corresponding values are \(y=-2\) and \(y=3\).

\(\{(3,-2),(-2,3)\}\)

- Every point on the unit circle is a distance \(1\) from the origin. - Use the Pythagorean relationship between \(x\) and \(y\). - Use the quadrant condition when choosing a square root. - A point lies on the circle only if its coordinates satisfy \(x^2+y^2=1\).

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.

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

- 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.

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 \([-1, 7]\).

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 \([-1, 7]\).

- Isolate \(y^2\) first, then use the sign restriction to choose one square-root branch. - What condition must the expression under the square root satisfy? - Compare \(g(-x)\) with \(g(x)\) to test y-axis symmetry. - What information is lost if you keep only one sign after taking a square root?

1. From \(x^2+y^2=64\), \(y^2=64-x^2\). Under the restriction \(y\le0\), \(g(x)=-\sqrt{64-x^2}\). 2. The radicand must be nonnegative: \(64-x^2\ge0\). Thus \(-8\le x\le8\), so the domain is \([-8,8]\). 3. Because \(g(x)\le0\), with \(g(0)=-8\) and \(g(\pm8)=0\), the range is \([-8,0]\). 4. Since \(g(-x)=-\sqrt{64-(-x)^2}=g(x)\), the graph is symmetric about the y-axis. 5. Solving the circle equation without the restriction gives \(y=\pm\sqrt{64-x^2}\). The function \(g\) uses only the negative branch, so it represents the lower semicircle, not the full circle.

a) \(g(x)=-\sqrt{64-x^2}\) b) Domain: \([-8,8]\); range: \([-8,0]\) c) Yes. Since \(g(-x)=g(x)\), the graph is symmetric about the y-axis. d) The full circle requires both \(y=\sqrt{64-x^2}\) and \(y=-\sqrt{64-x^2}\); \(g\) is only the lower semicircle.

- Substitute each point into the general circle equation. - Eliminate one unknown at a time by subtracting pairs of equations. - Complete the squares after determining the coefficients.

1. Substituting the three points gives \(2+D+E+F=0\), \(26+5D+E+F=0\), and \(26+D+5E+F=0\). 2. Subtracting the first equation from the second gives \(24+4D=0\), so \(D=-6\). 3. Subtracting the first equation from the third gives \(24+4E=0\), so \(E=-6\). 4. The first equation then gives \(F=10\). 5. Thus \(x^2+y^2-6x-6y+10=0\), which becomes \((x-3)^2+(y-3)^2=8\). 6. The center is \((3, 3)\), and the radius is \(2\sqrt{2}\).

\((x-3)^2+(y-3)^2=8\) Center: \((3, 3)\) Radius: \(2\sqrt{2}\)

- Complete the squares to reveal the center and the radius expression. - Compare the radius with the center's perpendicular distance to the x-axis. - Locate the tangency point along that perpendicular direction.

1. Complete the squares: \((x-3)^2+(y+2)^2=13-k\). 2. The center is \((3, -2)\), whose distance from the x-axis is \(2\). 3. Tangency to the x-axis requires the radius to equal that distance, so \(r=2\) and \(r^2=4\). 4. Therefore \(13-k=4\), giving \(k=9\). 5. The tangency point lies vertically above the center at \((3, 0)\).

\(k=9\) Center: \((3, -2)\) Radius: \(2\) Tangency point: \((3, 0)\)

- Draw horizontal and vertical legs from the center to a general point \(P(x,y)\). - Express each leg length using coordinate differences. - How is the distance from the center to every point on the circle related to \(r\)? - For part b, first determine \(r^2\) from the center and the given point.

1. From the center \((h,k)\) to \(P(x,y)\), the horizontal and vertical changes are \(x-h\) and \(y-k\). 2. These changes are the legs of a right triangle whose hypotenuse is the radius \(r\). By the Pythagorean theorem, \((x-h)^2+(y-k)^2=r^2\). 3. For the given circle, the squared radius is the squared distance from \((-3,2)\) to \((1,5)\): \(r^2=(1+3)^2+(5-2)^2=4^2+3^2=25\). 4. Substitute \(h=-3\), \(k=2\), and \(r^2=25\): \((x+3)^2+(y-2)^2=25\).

a) \((x-h)^2+(y-k)^2=r^2\) b) \((x+3)^2+(y-2)^2=25\)

- Group the x-terms and y-terms before completing either square. - Whatever constants you add to create perfect-square trinomials must also be added to the other side of the equation. - After finding standard form, substitute the line's fixed y-value and count the real intersection points.

1. Move the constant term: \(x^2-6x+y^2+8y=11\). 2. Complete each square by adding \(9\) and \(16\) to both sides: \((x-3)^2+(y+4)^2=11+9+16=36\). 3. Therefore, the center is \((3,-4)\) and the radius is \(6\). 4. Substitute \(y=2\): \((x-3)^2+(2+4)^2=36\), so \((x-3)^2=0\) and \(x=3\). 5. The line has exactly one intersection point, \((3,2)\), so \(y=2\) is tangent to the circle.

a) \((x-3)^2+(y+4)^2=36\) b) Center: \((3,-4)\); radius: \(6\) c) \(y=2\) is tangent at \((3,2)\).

- Solve the linear equation for \(x\). - Substitute and expand the squared binomial. - Factor the resulting quadratic in \(y\), then find the matching \(x\)-values.

1. Solve the linear equation for \(x\): \(x=y+c\). 2. Substitute into the first equation: \((y+c)^2+y^2=25c^2\). 3. Expand and simplify: \(2y^2+2cy-24c^2=0\). Divide by \(2\): \(y^2+cy-12c^2=0\). 4. Factor: \((y-3c)(y+4c)=0\). Thus, \(y=3c\) or \(y=-4c\). 5. Using \(x=y+c\), the ordered pairs are \((4c,3c)\) and \((-3c,-4c)\). 6. When \(c=0\), these expressions represent the same ordered pair, \((0,0)\).

a) \(y^2+cy-12c^2=0\) b) For \(c\ne0\), \(\{(4c,3c),(-3c,-4c)\}\). For \(c=0\), \(\{(0,0)\}\).

- Write the circle equation and identify the parabola's vertex. - Use the condition that the vertex lies on the circle. - Rewrite one equation so that \(x^2\) can be substituted into the other.

1. The circle has equation \(x^2+y^2=100\), and the parabola has vertex \((0,c)\). 2. For the vertex to lie on the circle, \(c^2=100\). Since \(c<0\), \(c=-10\). 3. The system is \(x^2+y^2=100\) and \(y=\frac{1}{4}x^2-10\). Rewrite the second equation as \(x^2=4y+40\). 4. Substitute into the circle equation: \(4y+40+y^2=100\), so \(y^2+4y-60=0\). 5. Factor: \((y+10)(y-6)=0\). Thus, \(y=-10\) or \(y=6\). 6. When \(y=-10\), \(x^2=0\), giving \((0,-10)\). When \(y=6\), \(x^2=64\), giving \((8,6)\) and \((-8,6)\).

a) \(c=-10\) b) \(\{(0,-10),(8,6),(-8,6)\}\)

- Solve the linear equation for one variable. - Substitute and simplify the resulting quadratic. - Factor, then check whether the two ordered-pair expressions coincide for a special value of \(k\).

1. Solve the first equation for \(y\): \(y=5k-x\). 2. Substitute into the second equation: \(x^2+(5k-x)^2=13k^2\). 3. Expand and simplify: \(2x^2-10kx+12k^2=0\). Divide by \(2\): \(x^2-5kx+6k^2=0\). 4. Factor: \((x-2k)(x-3k)=0\). Thus, \(x=2k\) or \(x=3k\). 5. Using \(y=5k-x\), the ordered pairs are \((2k,3k)\) and \((3k,2k)\). 6. When \(k=0\), these expressions represent the same ordered pair, \((0,0)\).

For \(k\ne0\), \(\{(2k,3k),(3k,2k)\}\). For \(k=0\), \(\{(0,0)\}\).

- Solve the linear equation for one variable. - Substitute and simplify the resulting quadratic. - After factoring, determine when the two ordered-pair expressions coincide.

1. Solve the linear equation for \(y\): \(y=5m-x\). 2. Substitute into the quadratic equation: \(x^2+(5m-x)^2=17m^2\). 3. Expand and simplify: \(2x^2-10mx+8m^2=0\). Divide by \(2\): \(x^2-5mx+4m^2=0\). 4. Factor: \((x-4m)(x-m)=0\). Thus, \(x=4m\) or \(x=m\). 5. The corresponding values of \(y\) are \(m\) and \(4m\). When \(m=0\), the two ordered-pair expressions coincide.

For \(m\ne0\), \(\{(4m,m),(m,4m)\}\). For \(m=0\), \(\{(0,0)\}\).

- If a circle in Quadrant I is tangent to both axes, how are its center coordinates related to its radius? - Write the standard circle equation using one unknown radius before using point \(P\). - Substitute the coordinates of \(P\) and simplify the resulting equation in the radius. - Check whether every algebraic radius you obtain is positive and satisfies the geometric conditions.

1. Because the circle is tangent to both coordinate axes and its center is in Quadrant I, if its radius is \(r>0\), then its center is \((r,r)\). 2. Since \(P(6,2)\) lies on the circle, \((6-r)^2+(2-r)^2=r^2\). 3. Expand and simplify: \(40-16r+2r^2=r^2\), so \(r^2-16r+40=0\). 4. Solve the quadratic: \(r=\frac{16\pm\sqrt{256-160}}{2}=8\pm2\sqrt{6}\). Both values are positive, so both satisfy the Quadrant I condition. 5. For each radius, the center is \((r,r)\). Therefore, the two circle equations are obtained by substituting \(r=8-2\sqrt{6}\) and \(r=8+2\sqrt{6}\) into \((x-r)^2+(y-r)^2=r^2\).

\((x-(8-2\sqrt{6}))^2+(y-(8-2\sqrt{6}))^2=(8-2\sqrt{6})^2\) \((x-(8+2\sqrt{6}))^2+(y-(8+2\sqrt{6}))^2=(8+2\sqrt{6})^2\) There are two solutions because the condition that \(P(6,2)\) lies on the tangent-to-both-axes circle produces two positive radius values.