Standard Form of Equation of Circle

[nycmath]

Precalculus
Michael Sullivan
Edition 10
Chapter 1, Section 1.4

Find the standard form of the equation of each circle.

38. Center (-3, 1) and tangent to the y-axis.

42. Center (4, -2) and tangent to the line x = 1.

[Math Tutor]

The trick with tangent lines is that the radius is just the perpendicular distance from the center to the line.

38. Center [tex](-3, 1)[/tex], tangent to the y-axis (which is the line [tex]x = 0[/tex]). The distance from the center to the y-axis is the absolute value of its x-coordinate:
[tex]r = |-3| = 3[/tex]
Answer: [tex](x + 3)^2 + (y - 1)^2 = 9[/tex]

42. Center [tex](4, -2)[/tex], tangent to the vertical line [tex]x = 1[/tex]. Same idea, just measure horizontally:
[tex]r = |4 - 1| = 3[/tex]
Answer: [tex](x - 4)^2 + (y + 2)^2 = 9[/tex]

If it helps, sketch it. The circle only touches the line at one point, so the shortest distance from center to line is exactly the radius. For a vertical line you only care about the x-values, and for a horizontal line you would only care about the y-values.

[nycmath]

Math Tutor,

Thank you for your reply and effort. Thank you for the tip to graph the circle for a clear understanding.

[Eigenvalue]

Someone already provided an answer, but I decided to solve this one as well:
The equation for a circle is [tex](x-h)^{2 }[/tex]+[tex](y-k)^{2 }[/tex]=[tex]r^{2 }[/tex], with (h,k) as the center and r as the radius.

1. Since the circle is tangent to the y-axis (x=0), then one of the coordinates on the circle will have an x-value equal to 0
The point is (-3, 0)
The center is (-3, 1)
We can use the distance formula to find the radius; r= [tex]\sqrt{(-3+3)^{2}+(1-0)^{2}}[/tex]
r=[tex]\sqrt{9}[/tex]
r=3
The equation of the circle is [tex](x+3)^{2 }[/tex]+[tex](y-1)^{2 }[/tex]=0

2. Since the circle is tangent to the line x=1, then one of the coordinates on the circle will have an x-value of 1
The point is (1, -2)
The center is (4, -2)
We can use the distance formula to find the radius; r=[tex]\sqrt{(1-4)^{2}+(-2+2)^{2}}[/tex]
r=[tex]\sqrt{9}[/tex]
r=3
The equation of the circle is [tex](x-4)^{2 }[/tex]+[tex](y+2)^{2 }[/tex]=9

[Eigenvalue]

Generalized formula given the center and a tangent line:
1. center (h,k) and tangent to y=c
Since the circle is tangent to y=c, one of the coordinates on the circle will have a y-coordinate of c
The point is (h,c)
We can use the distance formula to find the radius; r=[tex]\sqrt{(h-h)^{2}+(k-c)^{2}}[/tex]
r=[tex]\sqrt{(k-c)^{2}}[/tex]
The equation is [tex](x-h)^{2 }[/tex]+[tex](y-k)^{2 }[/tex]=[tex](k-c)^{2 }[/tex]

1. center (h,k) and tangent to x=c
Since the circle is tangent to x=c, one of the coordinates on the circle will have a x-coordinate of c
The point is (c,k)
We can use the distance formula to find the radius; r=[tex]\sqrt{(h-c)^{2}+(k-k)^{2}}[/tex]
r=[tex]\sqrt{(h-c)^{2}}[/tex]
The equation is [tex](x-h)^{2 }[/tex]+[tex](y-k)^{2 }[/tex]=[tex](h-c)^{2 }[/tex]

[nycmath]

Eigenvalue,

Thank you for your fabulous reply.