Question

for exercises 17 - 32, information about a circle is given. a. write an equation of the circle in standard form. b. graph the circle. (see examples 1 - 2) 17. center: (-2, 5); radius: 1 19. center: (-4, 1); radius: 3 21. center: (-4, -3); radius: $sqrt{11}$ 23. center: (0, 0); radius: 2.6 25. the endpoints of a diameter are (-2, 4) and (6, -2). 27. the center is (-2, -1) and a point on the circle is (6, 5). 29. the center is (4, 6) and the circle is tangent to the y - axis. (informally, a line is tangent to a circle if it touches the circle in exactly one point.) 31. the center is in quadrant iv, the radius is 5, and the circle is tangent to both the x - and y - axes. 33. write an equation that represents the set of points that are 5 units from (8, -11). 35. write an equation of the circle that is tangent to both axes with radius $sqrt{7}$ and center in quadrant i.

Answer:

1. For problem 17:

• a. Write the equation of the circle in standard - form:
• The standard form of a circle equation is \((x - h)^2+(y - k)^2=r^2\), where \((h,k)\) is the center of the circle and \(r\) is the radius.
• Given the center \((h,k)=(-2,5)\) and \(r = 1\).
• Substitute \(h=-2\), \(k = 5\), and \(r = 1\) into the standard - form equation:
• \((x-(-2))^2+(y - 5)^2=1^2\).
• Simplify to \((x + 2)^2+(y - 5)^2=1\).
• b. Graph the circle:
• The center of the circle is at the point \((-2,5)\) on the coordinate plane.
• Since the radius \(r = 1\), we can plot points 1 unit away from the center in all directions (up, down, left, and right) and then draw a smooth curve to form the circle.

1. For problem 19:

• a. Write the equation of the circle in standard - form:
• Given the center \((h,k)=(-4,1)\) and \(r = 3\).
• Substitute into the standard - form \((x - h)^2+(y - k)^2=r^2\):
• \((x-(-4))^2+(y - 1)^2=3^2\).
• Simplify to \((x + 4)^2+(y - 1)^2=9\).
• b. Graph the circle:
• The center is at \((-4,1)\).
• Since \(r = 3\), we can plot points 3 units away from the center in all directions and draw the circle.

1. For problem 21:

• a. Write the equation of the circle in standard - form:
• Given the center \((h,k)=(-4,-3)\) and \(r=\sqrt{11}\).
• Substitute into the standard - form \((x - h)^2+(y - k)^2=r^2\):
• \((x-(-4))^2+(y-(-3))^2=(\sqrt{11})^2\).
• Simplify to \((x + 4)^2+(y + 3)^2=11\).
• b. Graph the circle:
• The center is at \((-4,-3)\).
• Since \(r=\sqrt{11}\approx3.32\), we can estimate points approximately 3.32 units away from the center in all directions and draw the circle.

1. For problem 23:

• a. Write the equation of the circle in standard - form:
• Given the center \((h,k)=(0,0)\) and \(r = 2.6\).
• Substitute into the standard - form \((x - h)^2+(y - k)^2=r^2\):
• \((x-0)^2+(y - 0)^2=(2.6)^2\).
• Simplify to \(x^2+y^2 = 6.76\).
• b. Graph the circle:
• The center is at the origin \((0,0)\).
• Since \(r = 2.6\), we can plot points 2.6 units away from the origin in all directions and draw the circle.

1. For problem 25:

• First, find the center of the circle. The center of a circle is the mid - point of the diameter. The mid - point formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \((\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})\).
• Given the endpoints of the diameter \((x_1,y_1)=(-2,4)\) and \((x_2,y_2)=(6,-2)\).
• The center \((h,k)=(\frac{-2 + 6}{2},\frac{4+( - 2)}{2})=(2,1)\).
• The radius \(r\) is the distance from the center \((2,1)\) to either of the endpoints. Using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), with \((x_1,y_1)=(2,1)\) and \((x_2,y_2)=(-2,4)\):
• \(r=\sqrt{(-2 - 2)^2+(4 - 1)^2}=\sqrt{(-4)^2+3^2}=\sqrt{16 + 9}=\sqrt{25}=5\).
• The standard - form of the circle equation is \((x - 2)^2+(y - 1)^2=25\).
• b. Graph the circle:
• The center is at \((2,1)\).
• Since \(r = 5\), we can plot points 5 units away from the center in all directions and draw the circle.

1. For problem 27:

• First, find the radius. The radius \(r\) is the distance between the center \((h,k)=(-2,-1)\) and the point \((x,y)=(6,5)\) on the circle. Using the distance formula \(r=\sqrt{(x - h)^2+(y - k)^2}\):
• \(r=\sqrt{(6-(-2))^2+(5-(-1))^2}=\sqrt{(6 + 2)^2+(5 + 1)^2}=\sqrt{64+36}=\sqrt{100}=10\).
• The standard - form of the circle equation is \((x + 2)^2+(y + 1)^2=100\).
• b. Graph the circle:
• The center is at \((-2,-1)\).
• Since \(r = 10\), we can plot points 10 units away from the center in all directions and draw the circle.

1. For problem 29:

• Since the circle with center \((h,k)=(4,6)\) is tangent to the \(y\) - axis, the radius \(r\) is the \(x\) - coordinate of the center (the distance from the center to the \(y\) - axis). So \(r = 4\).
• The standard - form of the circle equation is \((x - 4)^2+(y - 6)^2=16\).
• b. Graph the circle:
• The center is at \((4,6)\).
• Since \(r = 4\), we can plot points 4 units away from the center in all directions and draw the circle.

1. For problem 31:

• Since the circle is in Quadrant IV and tangent to both the \(x\) - and \(y\) - axes and has a radius \(r = 5\), the center \((h,k)=(5,-5)\).
• The standard - form of the circle equation is \((x - 5)^2+(y + 5)^2=25\).
• b. Graph the circle:
• The center is at \((5,-5)\).
• Since \(r = 5\), we can plot points 5 units away from the center in all directions and draw the circle.

1. For problem 33:

• The set of points that are 5 units from the point \((h,k)=(8,-11)\) is a circle.
• The standard - form of the circle equation is \((x - 8)^2+(y + 11)^2=25\).

1. For problem 35:

• Since the circle is tangent to both axes and in Quadrant I with radius \(r=\sqrt{7}\), the center \((h,k)=(\sqrt{7},\sqrt{7})\).
• The standard - form of the circle equation is \((x-\sqrt{7})^2+(y - \sqrt{7})^2=7\).