College Algebra Questions and Problems With Solutions 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.
Questions
Question 1
Find the equation of a circle whose center is at the point \((-2 , 3)\) and its diameter has a length of \(10\).
Question 2
Find the center and the radius of the circle whose equation is given by \(x^2 + 4x + y^2 - 8y = 5\).
Question 3
Find an equation of the circle that is is tangent to both the \(x\) and \(y\) axes, with a radius of \(4\) and whose center is located in the second quadrant.
Question 4
Find an equation of the circle whose center is at the point \((-4 , 6)\) and passes through the point \((1 , 2)\).
Question 5
Find an equation of the circle whose center is at the point \((-3 , 6)\) and is tangent to the \(y\) axis.
Question 6
Find an equation of the circle whose center is at the point \((2 , -5)\) and is tangent to the \(x\) axis.
Question 7
Find an equation of the circle whose diameter has endpoints at \((-5 , 2)\) and \( (3 , 6)\).
Question 8
Find the \(x\) and \(y\) intercepts of the graph of the circle given by the equation
\(x^2 + 3x + y^2 - 4y = 18\)
Question 9
Is point A(-3 , 8) inside, outside or on the circle whose equation is given by
\(x^2 + 4x + y^2 - 8y = 5\)
Question 10
Find the points of intersection of the circle with equation
\((x - 2)^2 + (y - 6)^2 = 40\)
and the line with equation \(y = 3x\)
Question 11
Three points are defined as follows: A(-2 , 1), B(6 , 1) and C(-2 , 7).
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.
Solutions to the Above Questions
Solution to Question 1
Find the equation of a circle whose center is at the point \((-2 , 3)\) and its diameter has a length of \(10\).
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: \((x + 2)^2 + (y - 3)^2 = 25\)
Solution to Question 2
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.
\((x^2 + 4x + 4 - 4) + (y^2 - 8y + 16 - 16) = 5\)
\((x + 2)^2 + (y - 4)^2 = 25\).
Identify the above as the standard equation of a circle with center \((-2 , 4)\) and radius \(5\).
Solution to Question 3
Find an equation of the circle that is is tangent to both the \(x\) and \(y\) axes, with a radius of \(
4\) and whose center is located in the second quadrant.
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
\((x + 4)^2 + (y - 4)^2 = 16\)
Solution to Question 4
We first need to find the radius which is given by the distance from the center to the point \( (1 , 2)\)
\(r = \sqrt{(6 - 2)^2 + (-4 - 1)^2} = \sqrt{41}\)
The equation is given by
\((x + 4)^2 + (y - 6)^2 = 41\)
Solution to Question 5
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 and that is \(|-3| = 3\).
The equation is given by
\((x + 3)^2 + (y - 6)^2 = 9\)
Solution to Question 6
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 and that is \(|-5| = 5\).
The equation is given by
\((x - 2)^2 + (y + 5)^2 = 25\)
Solution to Question 7
We first find the radius \(r\) which is half the length of the diameter
\(r = \frac{1}{2} \sqrt{64 + 16} = 2 \sqrt{5}\)
The center \(C\) is the midpoint of \(A\) and \(B\): \(C(-1 , 4)\)
The equation is give by: \((x + 1)^2 + (y - 4)^2 = 20\)
Solution to Question 8
\(x\) intercept (set \(y = 0\) in given equation) : \(x^2 + 3x = 18\)
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 \((0 , 2 + \sqrt{22})\)
Solution to Question 9
We first write the equation in standard form and find the center and radius of the circle. By completing the squares, the given equation can be written as follows
\((x + 2)^2 + (y - 4)^2 = 5^2\)
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.
Solution to Question 10
We need to solve the system of equations as follows: Substitute \(y\) by \(3x\) in the equation of 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 \((4 , 12)\)
Solution to Question 11
a) \(M: (2 , 4)\)
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
\((x - 2)^2 + (y - 4)^2 = 25\)
