Equations of Circles: Practice Problems with Solutions

The standard equation of a circle centered at the origin (0, 0) is \(x^2+y^2=r^2,\) where \(r\) represents the radius. Given that the diameter is \(2\sqrt{11},\) the radius is half of that value:

\(r=\dfrac{2\sqrt{11}}{2}=\sqrt{11}\)

Substituting into the standard form:

\(x^2+y^2=(\sqrt{11})^2 \implies x^2+y^2=11\)

For a circle with center \((h, k)=(3,-4)\) and radius \(r=6,\) the standard equation is:

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

Substituting the given parameters:

\((x-3)^2+(y-(-4))^2=6^2\)

\((x-3)^2+(y+4)^2=36\)

With center at \((5,-12)\) and radius \(r=13,\) we apply standard form directly:

\((x-5)^2+(y-(-12))^2=13^2\)

Simplifying the squared terms:

\((x-5)^2+(y+12)^2=169\)

Given the center \((5,-3)\) and radius \(\sqrt{6},\) the equation yields:

\((x-5)^2+(y+3)^2=(\sqrt{6})^2\)

Squaring the radical on the right side:

\((x-5)^2+(y+3)^2=6\)

The circle with center \((h, k)=\left(\pi,-\dfrac{1}{12}\right)\) and radius \(r=\sqrt{17}\) has standard form:

\((x-\pi)^2+\left(y-\left(-\dfrac{1}{12}\right)\right)^2=(\sqrt{17})^2\)

Simplifying:

\((x-\pi)^2+\left(y+\dfrac{1}{12}\right)^2=17\)

For a circle centered at \((0, 8)\) with radius \(5,\) we set up the equation as:

\((x-0)^2+(y-8)^2=5^2\)

Simplifying the \(x\)-term and squaring the radius:

\(x^2+(y-8)^2=25\)

If the center is located at \((-1,-6)\) and the radius is \(8,\) we have:

\((x+1)^2+(y+6)^2=8^2\)

Evaluating the right side:

\((x+1)^2+(y+6)^2=64\)

For center \(\left(\dfrac{3}{5}, \dfrac{1}{3}\right)\) and radius \(\sqrt{10},\) the standard form is:

\(\left(x-\dfrac{3}{5}\right)^2+\left(y-\dfrac{1}{3}\right)^2=(\sqrt{10})^2\)

Simplifying the radius:

\(\left(x-\dfrac{3}{5}\right)^2+\left(y-\dfrac{1}{3}\right)^2=10\)

Given endpoints of a diameter at \(A(-1, 5)\) and \(B(-5, 7).\)

We find the center \((h, k)\) as the midpoint of segment \(AB:\)

\(h=\dfrac{-1+(-5)}{2}=-3, \quad k=\dfrac{5+7}{2}=6\)

The radius is half the length of diameter \(AB:\)

\(\text{Distance } AB=\sqrt{(-5-(-1))^2+(7-5)^2}=\sqrt{16+4}=\sqrt{20}=2\sqrt{5}\)

\(r=\dfrac{2\sqrt{5}}{2}=\sqrt{5}\)

Therefore, the standard equation is:

\((x+3)^2+(y-6)^2=5\)

Given endpoints of a diameter at \(A(2, 3)\) and \(B(-4, 5).\)

First, find the center by calculating the midpoint:

\(h=\dfrac{2+(-4)}{2}=-1, \quad k=\dfrac{3+5}{2}=4\)

The radius is half the distance between \(A\) and \(B:\)

\(\text{Distance } AB=\sqrt{(-4-2)^2+(5-3)^2}=\sqrt{36+4}=\sqrt{40}=2\sqrt{10}\)

\(r=\dfrac{2\sqrt{10}}{2}=\sqrt{10}\)

The standard equation is:

\((x+1)^2+(y-4)^2=10\)

Given: Center at \((1, -3)\) and the circle passes through \((-3, 5).\)

The radius equals the distance between the center and the given point on the circle:

\(r=\sqrt{(-3-1)^2+(5-(-3))^2}=\sqrt{16+64}=\sqrt{80}=4\sqrt{5}\)

Writing in standard form with \(r^2 = 80:\)

\((x-1)^2+(y+3)^2=80\)

Given: Center lies on the x-axis (meaning \(k=0\)) and passes through \(A(1, 3)\) and \(B(4, 6).\)

Because the center \((h, 0)\) is equidistant from points \(A\) and \(B,\) we set up the distance equality:

\(\sqrt{(1-h)^2+(3-0)^2}=\sqrt{(4-h)^2+(6-0)^2}\)

Squaring both sides and simplifying:

\((1-h)^2+9=(4-h)^2+36\)

\(1-2h+h^2+9=16-8h+h^2+36\)

\(-2h+10=-8h+52 \implies 6h=42 \implies h=7\)

The center is \((7, 0).\) We calculate the radius using point \(A(1, 3):\)

\(r=\sqrt{(1-7)^2+(3-0)^2}=\sqrt{36+9}=\sqrt{45}=3\sqrt{5}\)

The standard equation is:

\((x-7)^2+y^2=45\)

Given: Center lies on the y-axis (meaning \(h=0\)) and passes through \(A(2, 2)\) and \(B(6,-4).\)

Since the center \((0, k)\) is equidistant from \(A\) and \(B:\)

\(\sqrt{(2-0)^2+(2-k)^2}=\sqrt{(6-0)^2+(-4-k)^2}\)

Squaring both sides and expanding:

\(4+(2-k)^2=36+(-4-k)^2\)

\(4+4-4k+k^2=36+16+8k+k^2\)

\(8-4k=52+8k \implies -12k=44 \implies k=-\dfrac{11}{3}\)

The center is \(\left(0,-\dfrac{11}{3}\right).\) Using point \(A(2, 2)\) to determine \(r^2:\)

\(r^2=(2-0)^2+\left(2-\left(-\dfrac{11}{3}\right)\right)^2=4+\left(\dfrac{17}{3}\right)^2=4+\dfrac{289}{9}=\dfrac{325}{9}\)

Thus, the standard equation is:

\(x^2+\left(y+\dfrac{11}{3}\right)^2=\dfrac{325}{9}\)

Given: Center at \((5, 3)\) and tangent to the y-axis.

When a circle is tangent to the y-axis (line \(x=0\)), the radius is equal to the absolute value of the center's x-coordinate:

\(r=|h|=|5|=5\)

Substituting center \((5, 3)\) and radius \(r=5\) into standard form:

\((x-5)^2+(y-3)^2=5^2\)

\((x-5)^2+(y-3)^2=25\)

Given: Tangent to both coordinate axes, radius \(5,\) and located in the third quadrant.

In the third quadrant, both coordinates are negative. Since the circle is tangent to both axes, \(|h|=r\) and \(|k|=r:\)

\(h=-5, \quad k=-5\)

Substituting into standard form:

\((x-(-5))^2+(y-(-5))^2=5^2\)

\((x+5)^2+(y+5)^2=25\)

Given: Center at \((-3, -4)\) and tangent to the line \(3x+4y=16.\)

The radius is the perpendicular distance from the center to the tangent line \(3x+4y-16=0.\) Using the point-to-line distance formula:

\(r=\dfrac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}=\dfrac{|3(-3)+4(-4)-16|}{\sqrt{3^2+4^2}}=\dfrac{|-9-16-16|}{5}=\dfrac{41}{5}\)

With center \((-3, -4)\) and \(r=\dfrac{41}{5},\) the standard equation is:

\((x+3)^2+(y+4)^2=\left(\dfrac{41}{5}\right)^2\)

\((x+3)^2+(y+4)^2=\dfrac{1681}{25}\)

Given: Center at \((5, -5)\) and tangent to the line \(5x-12y=24.\)

We find the radius using the distance from \((5, -5)\) to \(5x-12y-24=0:\)

\(r=\dfrac{|5(5)-12(-5)-24|}{\sqrt{5^2+(-12)^2}}=\dfrac{|25+60-24|}{13}=\dfrac{61}{13}\)

Substituting into standard form:

\((x-5)^2+(y+5)^2=\left(\dfrac{61}{13}\right)^2\)

\((x-5)^2+(y+5)^2=\dfrac{3721}{169}\)

Given: Center lies on the x-axis and the circle is tangent to the line \(x-y=2\) at the point \((4, 2).\)

Because the center is on the x-axis, its coordinates are \((h, 0).\) The distance from \((h, 0)\) to the line \(x-y-2=0\) must equal the distance from \((h, 0)\) to the point of tangency \((4, 2):\)

\(\dfrac{|h-0-2|}{\sqrt{1^2+(-1)^2}}=\sqrt{(h-4)^2+(0-2)^2}\)

\(\dfrac{|h-2|}{\sqrt{2}}=\sqrt{(h-4)^2+4}\)

Squaring both sides:

\(\dfrac{(h-2)^2}{2}=(h-4)^2+4 \implies (h-2)^2=2(h-4)^2+8\)

Expanding and collecting like terms:

\(h^2-4h+4=2(h^2-8h+16)+8\)

\(h^2-4h+4=2h^2-16h+40 \implies h^2-12h+36=0\)

\((h-6)^2=0 \implies h=6\)

The center is \((6, 0).\) We calculate the radius using the point of tangency \((4, 2):\)

\(r=\sqrt{(6-4)^2+(0-2)^2}=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}\)

Thus, the standard equation is:

\((x-6)^2+y^2=8\)

Given: The circle is tangent to the line \(x+2y=3\) at \((-1, 2),\) and the center lies on the y-axis.

Because the center is on the y-axis, it has coordinates \((0, k).\) The distance from \((0, k)\) to \(x+2y-3=0\) equals the distance from \((0, k)\) to the tangency point \((-1, 2):\)

\(\dfrac{|0+2k-3|}{\sqrt{1^2+2^2}}=\sqrt{(0-(-1))^2+(k-2)^2}\)

\(\dfrac{|2k-3|}{\sqrt{5}}=\sqrt{1+(k-2)^2}\)

Squaring both sides and simplifying:

\(\dfrac{(2k-3)^2}{5}=1+(k-2)^2 \implies 4k^2-12k+9=5\left(1+k^2-4k+4\right)\)

\(4k^2-12k+9=5k^2-20k+25 \implies k^2-8k+16=0\)

\((k-4)^2=0 \implies k=4\)

The center is \((0, 4).\) The radius is:

\(r=\sqrt{(0+1)^2+(4-2)^2}=\sqrt{1+4}=\sqrt{5}\)

The standard equation is:

\(x^2+(y-4)^2=5\)

Given: Tangent to the line \(4x+3y=4\) at \((4, -4),\) with its center lying on the line \(x-y=7.\)

Since the center \((h, k)\) lies on \(x-y=7,\) we have \(h-k=7,\) which means \(h=k+7.\) The radius is both the distance from \((h, k)\) to \(4x+3y-4=0\) and the distance from \((h, k)\) to \((4, -4):\)

\(\dfrac{|4h+3k-4|}{\sqrt{4^2+3^2}}=\sqrt{(h-4)^2+(k+4)^2}\)

Substituting \(h=k+7:\)

\(\dfrac{|4(k+7)+3k-4|}{5}=\sqrt{(k+7-4)^2+(k+4)^2}\)

\(\dfrac{|7k+24|}{5}=\sqrt{(k+3)^2+(k+4)^2}\)

Squaring both sides:

\(\dfrac{49k^2+336k+576}{25}=2k^2+14k+25\)

\(49k^2+336k+576=50k^2+350k+625 \implies k^2+14k+49=0\)

\((k+7)^2=0 \implies k=-7\)

Since \(h=k+7,\) we find \(h=-7+7=0.\) The center is \((0, -7).\) We calculate the radius:

\(r=\sqrt{(0-4)^2+(-7+4)^2}=\sqrt{16+9}=5\)

The standard equation is:

\(x^2+(y+7)^2=25\)

Given equation: \(x^2+y^2=25.\) This matches the standard form of a circle centered at the origin \((h, k)=(0, 0).\) Comparing it with \((x-h)^2+(y-k)^2=r^2,\) we see that \(r^2=25,\) which means \(r=\sqrt{25}=5.\)

Key parameters:

• Center: (0, 0)
• Radius: 5

Given equation: \(x^2+y^2+6x-4x-12=0.\) Combining like terms gives \(x^2+y^2+2x-12=0.\) Completing the square yields:

\((x+1)^2+y^2=13\)

Writing this in standard form \((x-h)^2+(y-k)^2=r^2,\) we express \((x+1)^2\) as \((x-(-1))^2,\) giving \(h=-1.\) Because there is no linear \(y\)-term, \(k=0.\) Hence, the center is \((-1, 0).\) Since \(r^2=13,\) the radius is \(r=\sqrt{13} \approx 3.61.\)

Key parameters:

• Center: (-1, 0)
• Radius: \(\sqrt{13} \approx 3.61\)

Given equation: \(x^2+y^2+4x+12y+36=0.\) Completing the square for both variables:

\((x+2)^2+(y+6)^2=4\)

We read off the center directly as \((h, k)=(-2, -6).\) The radius squared is \(r^2=4,\) so \(r=2.\)

Key parameters:

• Center: (-2, -6)
• Radius: 2

Given equation: \(x^2+y^2-8x-2y+1=0.\) Completing the square:

\((x-4)^2+(y-1)^2=16\)

This is in standard form with center \((h, k)=(4, 1)\) and radius \(r=\sqrt{16}=4.\)

Key parameters:

• Center: (4, 1)
• Radius: 4

Given equation: \(x^2+y^2-10x+4y-7=0.\) Completing the square produces:

\((x-5)^2+(y+2)^2=36\)

The center is \((5, -2)\) and the radius is \(r=\sqrt{36}=6.\)

Key parameters:

• Center: (5, -2)
• Radius: 6

Given equation: \(x^2+y^2-10x-24y+25=0.\) Completing the square:

\((x-5)^2+(y-12)^2=144\)

The center is \((5, 12)\) and the radius is \(r=\sqrt{144}=12.\)

Key parameters:

• Center: (5, 12)
• Radius: 12

Given equation: \(2x^2+2y^2+12x-2y-3=0.\) Dividing by 2 and completing squares gives:

\((x+3)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{43}{4} \implies (x+3)^2+(y-0.5)^2=10.75\)

The center is \((-3, 0.5)\) and the radius is \(r=\sqrt{10.75} \approx 3.28.\)

Key parameters:

• Center: (-3, 0.5)
• Radius: \(\sqrt{10.75} \approx 3.28\)

Given equation: \(3x^2+3y^2-6x+5y=0.\) Dividing by 3 and completing the square:

\((x-1)^2+\left(y+\dfrac{5}{6}\right)^2=\dfrac{61}{36}\)

The center is \(\left(1, -\dfrac{5}{6}\right)\) and the radius is \(r=\sqrt{\dfrac{61}{36}}=\dfrac{\sqrt{61}}{6} \approx 1.30.\)

Key parameters:

• Center: \(\left(1, -\dfrac{5}{6}\right)\)
• Radius: \(\dfrac{\sqrt{61}}{6} \approx 1.30\)

Starting with \(x^2+y^2-2x+4y+5=0,\) complete the square for \(x\) and \(y:\)

For \(x:\) \(x^2-2x = (x-1)^2-1\)

For \(y:\) \(y^2+4y = (y+2)^2-4\)

Substituting into the original equation:

\((x-1)^2-1+(y+2)^2-4+5=0 \implies (x-1)^2+(y+2)^2=0\)

The sum of two real squares equals zero if and only if both expressions are zero: \(x-1=0\) and \(y+2=0.\) Thus, the equation represents a single point: \((1, -2).\) It is not a circle with positive radius, nor is it an empty graph.

Given \(x^2+y^2-6x+2y+36=0,\) completing the square yields:

\((x-3)^2-9+(y+1)^2-1+36=0 \implies (x-3)^2+(y+1)^2=-26\)

Because the sum of squares of real numbers cannot be negative, there are no real solutions. The equation has no graph (empty set).

Given \(x^2+y^2+4x-8y-5=0,\) completing the square produces:

\((x+2)^2-4+(y-4)^2-16-5=0 \implies (x+2)^2+(y-4)^2=25\)

Since the right side is strictly positive (\(r^2=25 > 0\)), this equation represents a circle with center \((-2, 4)\) and radius \(r=5.\)

For \(x^2+y^2+10y=0,\) we complete the square for \(y:\)

\(x^2+(y+5)^2-25=0 \implies x^2+(y+5)^2=25\)

The equation represents a circle with center \((0, -5)\) and radius \(r=5.\)

For \(x^2+y^2-x=0,\) completing the square for \(x\) gives:

\(\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}+y^2=0 \implies \left(x-\dfrac{1}{2}\right)^2+y^2=\dfrac{1}{4}\)

This equation represents a circle with center \(\left(\dfrac{1}{2}, 0\right)\) and radius \(r=\dfrac{1}{2}.\)

Given \(x^2+y^2+8x+15=0,\) completing the square on \(x:\)

\((x+4)^2-16+y^2+15=0 \implies (x+4)^2+y^2=1\)

The equation represents a circle centered at \((-4, 0)\) with radius \(r=1.\)

Given \(x^2+y^2-4x-4y+9=0,\) completing the square on both variables:

\((x-2)^2-4+(y-2)^2-4+9=0 \implies (x-2)^2+(y-2)^2=-1\)

The right side is negative, which is impossible for the sum of two real squares. Hence, the equation has no graph.

For \(x^2+y^2+7x+5y+16=0,\) completing squares:

\(\left(x+\dfrac{7}{2}\right)^2-\dfrac{49}{4}+\left(y+\dfrac{5}{2}\right)^2-\dfrac{25}{4}+16=0 \implies \left(x+\dfrac{7}{2}\right)^2+\left(y+\dfrac{5}{2}\right)^2=\dfrac{5}{2}\)

Because \(\dfrac{5}{2} > 0,\) this represents a circle with center \(\left(-\dfrac{7}{2}, -\dfrac{5}{2}\right)\) and radius \(r=\sqrt{\dfrac{5}{2}}=\dfrac{\sqrt{10}}{2}.\)

Combine like terms in \(x:\) \(x^2+y^2+2x-12=0.\) Move the constant to the right and group the \(x\)-terms:

\((x^2+2x)+y^2=12\)

To complete the square for \(x^2+2x,\) take half of 2, which is 1, and square it (\(1^2=1\)). Add 1 to both sides:

\((x^2+2x+1)+y^2=12+1\)

Factoring into a binomial squared gives the standard form:

\((x+1)^2+y^2=13\)

From \(x^2+y^2+4x+12y+36=0,\) group terms and isolate the constant:

\((x^2+4x)+(y^2+12y)=-36\)

Complete the square for each group by adding \(\left(\dfrac{4}{2}\right)^2=4\) and \(\left(\dfrac{12}{2}\right)^2=36\) to both sides:

\((x^2+4x+4)+(y^2+12y+36)=-36+4+36\)

Factoring into perfect squares gives standard form:

\((x+2)^2+(y+6)^2=4\)

Given \(x^2+y^2-8x-2y+1=0,\) regroup terms:

\((x^2-8x)+(y^2-2y)=-1\)

Add \((-4)^2=16\) and \((-1)^2=1\) to both sides:

\((x^2-8x+16)+(y^2-2y+1)=-1+16+1\)

Factoring produces the standard form:

\((x-4)^2+(y-1)^2=16\)

Starting with \(x^2+y^2-10x+4y-7=0:\)

\((x^2-10x)+(y^2+4y)=7\)

Add \((-5)^2=25\) and \(2^2=4\) to both sides:

\((x^2-10x+25)+(y^2+4y+4)=7+25+4\)

The standard form is:

\((x-5)^2+(y+2)^2=36\)

Given \(x^2+y^2-10x-24y+25=0:\)

\((x^2-10x)+(y^2-24y)=-25\)

Complete the squares by adding \(25\) and \(144\) to both sides:

\((x^2-10x+25)+(y^2-24y+144)=-25+25+144\)

Standard form:

\((x-5)^2+(y-12)^2=144\)

For \(x^2+y^2+4x+12y+3=0:\)

\((x^2+4x)+(y^2+12y)=-3\)

Adding \(4\) and \(36\) to both sides:

\((x^2+4x+4)+(y^2+12y+36)=-3+4+36\)

Standard form:

\((x+2)^2+(y+6)^2=37\)

From \(x^2+y^2-3x-4y=0,\) group terms:

\((x^2-3x)+(y^2-4y)=0\)

Add \(\left(-\dfrac{3}{2}\right)^2=2.25\) and \((-2)^2=4\) to both sides:

\((x^2-3x+2.25)+(y^2-4y+4)=0+2.25+4\)

Standard form:

\((x-1.5)^2+(y-2)^2=6.25 \quad \text{or} \quad \left(x-\dfrac{3}{2}\right)^2+(y-2)^2=\dfrac{25}{4}\)

Given \(2x^2+2y^2+12x-2y-3=0,\) divide the entire equation by 2 first:

\(x^2+y^2+6x-y=\dfrac{3}{2}\)

Group terms and complete the squares by adding \(3^2=9\) and \((-0.5)^2=0.25:\)

\((x^2+6x+9)+(y^2-y+0.25)=1.5+9+0.25\)

Standard form:

\((x+3)^2+(y-0.5)^2=10.75 \quad \text{or} \quad (x+3)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{43}{4}\)

Given \(3x^2+3y^2-6x+5y=0,\) divide by 3:

\(x^2+y^2-2x+\dfrac{5}{3}y=0\)

Group terms and add \(1^2=1\) and \(\left(\dfrac{5}{6}\right)^2=\dfrac{25}{36}\) to both sides:

\((x^2-2x+1)+\left(y^2+\dfrac{5}{3}y+\dfrac{25}{36}\right)=0+1+\dfrac{25}{36}\)

Standard form:

\((x-1)^2+\left(y+\dfrac{5}{6}\right)^2=\dfrac{61}{36}\)

From \((x-3)^2+(y+2)^2=9,\) expand each squared binomial:

\(x^2-6x+9+y^2+4y+4=9\)

Collect terms on the left side:

\(x^2+y^2-6x+4y+(13-9)=0 \implies x^2+y^2-6x+4y+4=0\)

From \((x+1)^2+y^2=\dfrac{25}{4},\) expand the first binomial:

\(x^2+2x+1+y^2=\dfrac{25}{4}\)

Multiply the entire equation by 4 to eliminate denominators:

\(4x^2+8x+4+4y^2=25 \implies 4x^2+4y^2+8x-21=0\)

In normalized form (dividing by 4): \(x^2+y^2+2x-\dfrac{21}{4}=0.\)

Starting with \((x-4)^2+(y+3)^2=\dfrac{1}{9},\) expand the squared binomials:

\(x^2-8x+16+y^2+6y+9=\dfrac{1}{9}\)

Multiply through by 9:

\(9x^2-72x+144+9y^2+54y+81=1 \implies 9x^2+9y^2-72x+54y+224=0\)

In normalized form: \(x^2+y^2-8x+6y+\dfrac{224}{9}=0.\)

From \(x^2+y^2=25,\) we have \(r^2=25,\) so \(r=5.\)

Using the circumference formula \(C=2\pi r:\)

\(C=2\pi(5)=10\pi \text{ units}\)

Using the area formula \(A=\pi r^2:\)

\(A=\pi(5^2)=25\pi \text{ square units}\)

Completing the square on \(x^2+y^2+6x-4x-12=0\) gives \((x+1)^2+y^2=13,\) which gives \(r=\sqrt{13}.\)

The circumference is:

\(C=2\pi\sqrt{13}=2\sqrt{13}\pi \text{ units}\)

The area is:

\(A=\pi(\sqrt{13})^2=13\pi \text{ square units}\)

Completing the square on \(x^2+y^2+4x+12y+36=0\) gives \((x+2)^2+(y+6)^2=4,\) so \(r=2.\)

Circumference:

\(C=2\pi(2)=4\pi \text{ units}\)

Area:

\(A=\pi(2^2)=4\pi \text{ square units}\)

Completing the square on \(x^2+y^2-8x-2y+1=0\) yields \((x-4)^2+(y-1)^2=16,\) so \(r=4.\)

Circumference:

\(C=2\pi(4)=8\pi \text{ units}\)

Area:

\(A=\pi(4^2)=16\pi \text{ square units}\)

Completing the square on \(x^2+y^2-10x+4y-7=0\) yields \((x-5)^2+(y+2)^2=36,\) so \(r=6.\)

Circumference:

\(C=2\pi(6)=12\pi \text{ units}\)

Area:

\(A=\pi(6^2)=36\pi \text{ square units}\)

Each complete rotation covers a distance equal to the wheel's circumference \(C:\)

\(C=\pi d = 70\pi \approx 219.91 \text{ cm}\)

For 50 full rotations, the total distance traveled is:

\(\text{Total distance}=50 \times 219.91 \approx 10995.5 \text{ cm}\)

Converting centimeters to meters:

\(\dfrac{10995.5}{100} = 109.955 \approx 109.96 \text{ m}\)

The bicycle travels approximately 109.96 meters.

The table has a radius of \(r_{\text{table}} = \dfrac{1.5}{2} = 0.75\text{ m}.\) The overhang is \(20\text{ cm} = 0.2\text{ m},\) so the radius of the tablecloth is:

\(r_{\text{cloth}} = 0.75 + 0.2 = 0.95 \text{ m}\)

Calculating the circular area:

\(A = \pi r^2 = \pi (0.95)^2 = 0.9025\pi \approx 2.835 \text{ m}^2\)

Approximately 2.835 square meters of fabric are needed for each tablecloth.

Using the area formula \(A=\pi r^2:\)

\(r^2 = \dfrac{A}{\pi} = \dfrac{15.7}{\pi} \approx \dfrac{15.7}{3.1416} \approx 4.9975\)

Taking the square root:

\(r = \sqrt{4.9975} \approx 2.24 \text{ m}\)

The radius of the billboard is approximately 2.24 meters.

The area of the cover is:

\(A = \pi r^2 = \pi (0.3)^2 = 0.09\pi \approx 0.2827 \text{ m}^2\)

Applying two full coats doubles the coverage required:

\(A_{\text{total}} = 2 \times 0.2827 \approx 0.5655 \text{ m}^2\)

Since 1 liter covers \(3 \text{ m}^2:\)

\(\text{Paint needed} = \dfrac{0.5655}{3} \approx 0.1885 \text{ L}\)

Approximately 0.19 liters of paint are required.

With an initial radius \(r_1 = 12\text{ m}:\)

\(A_1 = \pi (12)^2 = 144\pi \text{ m}^2\)

When the radius is reduced to \(r_2 = 10\text{ m}:\)

\(A_2 = \pi (10)^2 = 100\pi \text{ m}^2\)

The reduction in coverage is the difference between both areas:

\(\Delta A = 144\pi - 100\pi = 44\pi \approx 138.23 \text{ m}^2\)

Answers:

• Total coverage at \(r = 12\text{ m}:\) \(144\pi \approx 452.39 \text{ m}^2\)
• Area reduction at \(r = 10\text{ m}:\) \(44\pi \approx 138.23 \text{ m}^2\)

The inner radius (fountain) is \(r = \dfrac{6}{2} = 3\text{ m},\) and the outer radius (walkway edge) is \(R = \dfrac{12}{2} = 6\text{ m}.\)

The walkable area forms an annulus (circular ring):

\(A = \pi R^2 - \pi r^2 = \pi (6^2 - 3^2) = \pi (36 - 9) = 27\pi \approx 84.82 \text{ m}^2\)

The area available for walking is \(27\pi \approx \mathbf{84.82\text{ square meters}}.\)

Place tower \(A\) at the origin \((0, 0)\) with signal boundary \(x^2+y^2=100,\) and tower \(B\) at \((15, 0)\) with boundary \((x-15)^2+y^2=100.\)

The distance between the two centers is \(d = 15\text{ km}.\) The sum of their coverage radii is \(r_1 + r_2 = 10 + 10 = 20\text{ km}.\)

Because \(d < r_1 + r_2\) (\(15 < 20\)), their circular ranges intersect and overlap by \(5\text{ km}\) along the straight line connecting them. The midpoint between the towers is \(7.5\text{ km}\) from each tower, which is within the \(10\text{ km}\) broadcast limit of both towers.

Therefore, there is no dead zone between the two towers.

The orbital radius is the sum of the Earth's radius and the satellite's altitude:

\(r = 6378 + 600 = 6978 \text{ km}\)

The distance traveled during one complete circular orbit equals the circumference:

\(C = 2\pi r = 2\pi (6978) \approx 2 \times 3.14159 \times 6978 \approx 43844.1 \text{ km}\)

The orbital path length is approximately 43,844.1 kilometers.