sample 7 : Equation of Circle

College algebra questions and problems on the equation of circle are presented. The solutions are at the bottom of the page. Helpful tutorials on circles is included in this site.

\(x^2 + 3x + y^2 - 4y = 18\)

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

\((x - 2)^2 + (y - 6)^2 = 40\)

and the line with equation \(y = 3x\)

a) Find the midpoint M of segment BC.

b) Find the lengths of the line segments MA, MB and MC.

c) Find the equation of the circle that passes through the three points A, B and C.

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 \(r = \frac{10}{2} = 5\) and \(h = -2\) and \(k = 3\).

The equation of this circle is given by: \((x + 2)^2 + (y - 3)^2 = 25\)

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

Complete the squares inside the brackets.

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

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

Identify the above as the standard equation of a circle with center \((-2 , 4)\) and radius \(5\).

Since the circle is tangent to the \(x\) axis, the distance from the center to \(x\) axis is equal to the radius \(4\). In the same way the circle is tangent to the \(y\) axis and therefore the distance from the center to the \(y\) axis is also equal to the radius. Hence the center has coordinates \((-4 , 4)\). The equation of this circle is given by

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

\(r = \sqrt{(6 - 2)^2 + (-4 - 1)^2} = \sqrt{41}\)

The equation is given by

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

The equation is given by

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

The equation is given by

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

\(r = \frac{1}{2} \sqrt{64 + 16} = 2 \sqrt{5}\)

The center \(C\) is the midpoint of \(A\) and \(B\): \(C(-1 , 4)\)

The equation is give by: \((x + 1)^2 + (y - 4)^2 = 20\)

solve for \(x\): \(x = 3\) and \(x = -6\) , 2 \(x\) intercepts: \((3 , 0)\) and \((-6 , 0)\)

\(y\) intercept (set \(x = 0\) in given equation) : \(y^2 - 4y = 18\)

solve for \(y\): \(y = 2 - \sqrt{22}\) and \(y = 2 + \sqrt{22}\) , 2 \(y\) intercepts: \((0 , 2 - \sqrt{22})\) and \((0 , 2 + \sqrt{22})\)

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

Center at \((-2 , 4)\) and radius \(= 5\)

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

\(d = \sqrt{1 + 16} = \sqrt{17} = 4.1\) (to the nearest tenth)

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

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

Solve for \(x\) to find the solutions: \(x = 0\) and \(x = 4\)

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

For \(x = 0\) , \(y = 3x = 0\) and for \(x = 4\) , \(y = 12\).

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

b) \(L(MA) = \sqrt{16 + 9} = 5\)

\(L(MB) = \sqrt{16 + 9} = 5\)

\(L(MC) = \sqrt{16 + 9} = 5\)

c) The three points are at equal distances from point \(M\) and therefore point \(M\) is the center of the circle; hence the equation is given by

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

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