College Algebra Questions and Problems With Solutions
Equation of Circle

The standard equation of a circle with radius \(r\) and center at the point \((h , k)\) is given by:

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

In this problem, the radius is \(r = \frac{10}{2} = 5\), with \(h = -2\) and \(k = 3\).

Substituting these values into the standard form gives:

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

Group the terms in \(x\) and \(x^2\) and the terms in \(y\) and \(y^2\) in the given equation:

\[(x^2 + 4x) + (y^2 - 8y) = 5\]

Complete the squares inside the brackets:

\[(x^2 + 4x + 4 - 4) + (y^2 - 8y + 16 - 16) = 5\] \[(x + 2)^2 - 4 + (y - 4)^2 - 16 = 5\] \[(x + 2)^2 + (y - 4)^2 = 5 + 4 + 16 = 25\]

Comparing this to the standard equation \((x - h)^2 + (y - k)^2 = r^2\), we identify the center at \((-2 , 4)\) and radius \(r = \sqrt{25} = 5\).

Since the circle is tangent to the \(x\)-axis, the distance from the center to the \(x\)-axis is equal to the radius \(4\). Similarly, being tangent to the \(y\)-axis means the distance from the center to the \(y\)-axis is also equal to \(4\).

Since the center is located in the second quadrant, its coordinates must be \((-4 , 4)\).

The equation of this circle is given by:

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

We first need to find the radius \(r\), which is given by the distance from the center \((-4, 6)\) to the point \((1, 2)\):

\[r = \sqrt{(1 - (-4))^2 + (2 - 6)^2} = \sqrt{5^2 + (-4)^2} = \sqrt{25 + 16} = \sqrt{41}\]

The equation of the circle is given by:

\[(x + 4)^2 + (y - 6)^2 = 41\]

Since the circle is tangent to the \(y\)-axis, the radius is equal to the distance from the center to the \(y\)-axis, which is the absolute value of the \(x\)-coordinate of the center:

\[r = |-3| = 3\]

The equation is given by:

\[(x + 3)^2 + (y - 6)^2 = 9\]

Since the circle is tangent to the \(x\)-axis, the radius is equal to the distance from the center to the \(x\)-axis, which is the absolute value of the \(y\)-coordinate of the center:

\[r = |-5| = 5\]

The equation is given by:

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

The center \(C\) is the midpoint of the two endpoints:

\[C = \left(\frac{-5 + 3}{2}, \frac{2 + 6}{2}\right) = (-1, 4)\]

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

\[r = \frac{1}{2} \sqrt{(3 - (-5))^2 + (6 - 2)^2} = \frac{1}{2} \sqrt{8^2 + 4^2} = \frac{1}{2} \sqrt{64 + 16} = \frac{1}{2} \sqrt{80} = 2\sqrt{5}\]

Thus, \(r^2 = (2\sqrt{5})^2 = 20\). The equation is given by:

\[(x + 1)^2 + (y - 4)^2 = 20\]

• \(x\)-intercepts: Set \(y = 0\) in the given equation: \[x^2 + 3x = 18 \implies x^2 + 3x - 18 = 0 \implies (x + 6)(x - 3) = 0\] Solving for \(x\): \(x = 3\) and \(x = -6\). The two \(x\)-intercepts are \((3 , 0)\) and \((-6 , 0)\).
• \(y\)-intercepts: Set \(x = 0\) in the given equation: \[y^2 - 4y = 18 \implies y^2 - 4y - 18 = 0\] Using the quadratic formula for \(y\): \[y = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(-18)}}{2} = \frac{4 \pm \sqrt{16 + 72}}{2} = \frac{4 \pm \sqrt{88}}{2} = 2 \pm \sqrt{22}\] The two \(y\)-intercepts are \((0 , 2 - \sqrt{22})\) and \((0 , 2 + \sqrt{22})\).

We first write the equation in standard form by completing the squares:

\[(x^2 + 4x + 4) + (y^2 - 8y + 16) = 5 + 4 + 16\] \[(x + 2)^2 + (y - 4)^2 = 25 = 5^2\]

The center is at \((-2 , 4)\) and the radius is \(r = 5\).

We now calculate the distance \(d\) from the center to point A\((-3 , 8)\):

\[d = \sqrt{(-3 - (-2))^2 + (8 - 4)^2} = \sqrt{(-1)^2 + 4^2} = \sqrt{1 + 16} = \sqrt{17} \approx 4.1\]

Since the distance from the center to point A is smaller than the radius (\(4.1 < 5\)), point A is inside the circle.

We need to solve the system of equations by substituting \(y = 3x\) into the equation of the circle:

\[(x - 2)^2 + (3x - 6)^2 = 40\]

Factor out 3 from the second term: \((3x - 6) = 3(x - 2)\), so \((3x - 6)^2 = 9(x - 2)^2\):

\[(x - 2)^2 + 9(x - 2)^2 = 40 \implies 10(x - 2)^2 = 40 \implies (x - 2)^2 = 4\]

Taking the square root gives \(x - 2 = \pm 2\), resulting in \(x = 0\) and \(x = 4\).

Use \(y = 3x\) to find the corresponding values of \(y\):

• For \(x = 0\), \(y = 3(0) = 0\)
• For \(x = 4\), \(y = 3(4) = 12\)

The points of intersection are \((0 , 0)\) and \((4 , 12)\).

• a) Midpoint M of BC: \[M = \left(\frac{6 - 2}{2}, \frac{1 + 7}{2}\right) = (2, 4)\]
• b) Lengths of line segments:
  \(L(MA) = \sqrt{(-2 - 2)^2 + (1 - 4)^2} = \sqrt{(-4)^2 + (-3)^2} = \sqrt{16 + 9} = 5\)
  \(L(MB) = \sqrt{(6 - 2)^2 + (1 - 4)^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = 5\)
  \(L(MC) = \sqrt{(-2 - 2)^2 + (7 - 4)^2} = \sqrt{(-4)^2 + 3^2} = \sqrt{16 + 9} = 5\)
• c) Equation of the circle:

  The three points are at equal distances from point \(M\), and therefore point \(M\) is the center of the circumscribed circle with radius \(r = 5\). The equation is given by:

  \[(x - 2)^2 + (y - 4)^2 = 25\]