Tutorials with detailed solutions to examples and matched exercises on finding the equation of a circle, radius, and center[cite: 1]. Detailed explanations are also provided. Each solution is hidden inside a collapsible dropdown so you can practice independently before reviewing the steps.
A circle is the set of points equidistant from a point \( C(h,k) \) called the center. The fixed distance \( r \) from the center to any point on the circle is called the radius.
The standard equation of a circle with center at \( C(h,k) \) and radius \( r \) is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Find the equation of a circle whose center is at \((2, -4)\) and radius \(5\).
Substitute \((h, k)\) by \((2, -4)\) and \(r\) by \(5\) in the standard equation to obtain:
\[ (x - 2)^2 + (y - (-4))^2 = 5^2 \]
Simplify:
\[ (x - 2)^2 + (y + 4)^2 = 25 \]
Set \(h\), \(k\), and \(r\) parameters into this applet and plot the circle. Verify graphically that the equation is that of the circle with the given center and radius.
Find the equation of a circle whose center is at \((2, -4)\) and radius \(3\).
Substitute \((h, k) = (2, -4)\) and \(r = 3\) into the standard equation:
\[ (x - 2)^2 + (y + 4)^2 = 3^2 \implies (x - 2)^2 + (y + 4)^2 = 9 \]
Find the equation of a circle that has a diameter with the endpoints given by the points \(A(-1, 2)\) and \(B(3, 2)\).
The center \(C\) of the circle is the midpoint of the line segment making the diameter \(AB\). We first use the midpoint formula to find the coordinates of \(C\):
\[ C\left( \frac{-1 + 3}{2}, \frac{2 + 2}{2} \right) = C(1,2) \]
The radius \(r\) is half the distance between \(A\) and \(B\). Hence:
\[ r = \frac{1}{2} \sqrt{ [3 - (-1)]^2 + [2 - 2]^2 } \]
\[ = \frac{1}{2} \sqrt{4^2 + 0^2} = 2 \]
The coordinates of \(C\) and the radius \(r\) are used in the standard equation of the circle to obtain the equation:
\[ (x - 1)^2 + (y - 2)^2 = 2^2 \implies (x - 1)^2 + (y - 2)^2 = 4 \]
Set the \(h\), \(k\) and \(r\) parameters into this applet and plot the circle.
Find the equation of a circle that has a diameter with the endpoints given by \(A(0, -2)\) and \(B(0, 2)\).
Center is at the midpoint of \(A\) and \(B\): \( C(0, 0) \)
Radius is half the distance between \(A\) and \(B\): \( r = \frac{1}{2}\sqrt{(0-0)^2 + (2 - (-2))^2} = 2 \)
Equation: \( x^2 + y^2 = 4 \)
Find the center and radius of the circle with equation:
\[ x^2 - 4x + y^2 - 6y + 9 = 0 \]
Group terms and complete the square:
\[ (x^2 - 4x + 4) - 4 + (y^2 - 6y + 9) - 9 + 9 = 0 \]
\[ (x - 2)^2 + (y - 3)^2 = 4 = 2^2 \]
Center at \(C(2, 3)\) and radius \(r = 2\).
Find the center and radius of the circle with equation:
\[ x^2 - 2x + y^2 - 8y + 1 = 0 \]
Group terms and complete the squares:
\[ (x^2 - 2x + 1) - 1 + (y^2 - 8y + 16) - 16 + 1 = 0 \]
\[ (x - 1)^2 + (y - 4)^2 - 16 = 0 \implies (x - 1)^2 + (y - 4)^2 = 4^2 \]
Radius = 4
Center at \( (1 , 4) \).
Is the point \(P(3, 4)\) inside, outside or on the circle with equation:
\[ (x + 2)^2 + (y - 3)^2 = 9 \]
Center \(C(-2, 3)\) and radius \(r = 3\). Distance from \(C\) to \(P(3, 4)\):
\[ \sqrt{[3 - (-2)]^2 + [4 - 3]^2} = \sqrt{5^2 + 1^2} = \sqrt{26} \approx 5.1 \]
Since \(\sqrt{26} > 3\), point \(P\) is outside the circle.
Is the point \(P(-1, -3)\) inside, outside or on the circle with equation:
\[ (x - 1)^2 + (y + 3)^2 = 4 \]
Center is at \( (1 , -3) \) and radius is \(r = 2\).
Distance \( d \) from center to the point \((-1 , -3)\):
\[
d = \sqrt{(-1 - 1)^2 + (-3 + 3)^2} = \sqrt{(-2)^2 + 0} = 2
\]
Since \(d = r = 2\), point P is on the circle.
Find the equation of the circle such that the three points \(A(0, 4)\), \(B(3, 5)\) and \(D(7, 3)\) are on the circle.
Setting distances from center \(C(h,k)\) to \(A, B, D\) equal to radius \(r\) yields the system:
\[ \begin{cases} 2k + 6h = 18 \\ -2k + 14h = 42 \end{cases} \]
Solving gives \(h = 3\) and \(k = 0\). The radius is \(r = \sqrt{(3-0)^2 + (0-4)^2} = 5\).
Equation: \((x - 3)^2 + y^2 = 25\).
Find the equation of the circle such that the three points \(A(-5, 0)\), \(B(1, 0)\) and \(D(-2, -3)\) are on the circle.
Let the center be \(C(h,k)\). Equating the squared distances from \(C\) to \(A(-5,0)\), \(B(1,0)\), and \(D(-2,-3)\):
\[ (h + 5)^2 + k^2 = (h - 1)^2 + k^2 \implies h^2 + 10h + 25 = h^2 - 2h + 1 \implies 12h = -24 \implies h = -2 \]
Now equate distances from \(B\) and \(D\):
\[ (-2 - 1)^2 + k^2 = (-2 + 2)^2 + (k + 3)^2 \implies 9 + k^2 = (k + 3)^2 \implies 6k = 0 \implies k = 0 \]
Center is \(C(-2, 0)\). Find radius squared using point \(B(1,0)\):
\[ r^2 = (1 - (-2))^2 + (0 - 0)^2 = 3^2 = 9 \]
Equation of the circle:
\[ (x + 2)^2 + y^2 = 9 \]
Find the equation of the circle that is tangent to the line whose equation is given by \(x + y = 2\) and has its center at \((3, 5)\).
The perpendicular line through center \((3,5)\) to \(x + y = 2\) ($m = -1$) has slope \(m_2 = 1\) and equation \(y = x + 2\). Intersection (point of tangency) is at \((0,2)\).
Radius \(r = \sqrt{(3-0)^2 + (5-2)^2} = 3\sqrt{2}\).
Equation: \((x - 3)^2 + (y - 5)^2 = 18\).
Find the equation of the circle that is tangent to the line whose equation is given by \(x + 2y = 2\) and has its center at \((0,5)\).
The given line is \(x + 2y = 2\), which can be written as \(y = -\frac{1}{2}x + 1\), with slope \(m_1 = -\frac{1}{2}\).
The line passing through the center \((0, 5)\) perpendicular to the tangent line has slope \(m_2 = -\frac{1}{m_1} = 2\). Its equation is:
\[ y - 5 = 2(x - 0) \implies y = 2x + 5 \]
Find the point of tangency by substituting \(y = 2x + 5\) into the tangent line equation \(x + 2y = 2\):
\[ x + 2(2x + 5) = 2 \implies 5x + 10 = 2 \implies 5x = -8 \implies x = -\frac{8}{5} \]
Corresponding \(y\)-coordinate:
\[ y = 2\left(-\frac{8}{5}\right) + 5 = -\frac{16}{5} + \frac{25}{5} = \frac{9}{5} \]
The radius squared \(r^2\) is the squared distance from center \((0,5)\) to the point of tangency \(\left(-\frac{8}{5}, \frac{9}{5}\right)\):
\[ r^2 = \left(-\frac{8}{5} - 0\right)^2 + \left(\frac{9}{5} - 5\right)^2 = \frac{64}{25} + \left(-\frac{16}{5}\right)^2 = \frac{64}{25} + \frac{256}{25} = \frac{320}{25} = \frac{64}{5} \]
Equation of the circle:
\[ x^2 + (y - 5)^2 = \frac{64}{5} \]