
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 = 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 = √[ (6  2)^{ 2} + (4  1)^{ 2} ] = √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 = 1/2 √(64 + 16) = 2 √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  √22 and y = 2 + √22 , 2 y intercepts: (0 , 2  √22) and (0 , 2 + √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 = √(1 + 16) = √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) = √(16 + 9) = 5
L(MB) = √(16 + 9) = 5
L(MC) = √(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
More References and links
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