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13. what is the standard form equation of a circle with center at (3, -…

Question

  1. what is the standard form equation of a circle with center at (3, - 2) and radius 4? a. (x - 3)^2+(y + 2)^2 = 8 b. (x + 3)^2+(y + 2)^2 = 8 c. (x - 3)^2+(y + 2)^2 = 16 d. (x + 3)^2+(y - 2)^2 = 16 14. which of the following is the equation of a circle with center (0, - 1) and radius 6? a. x^2+y^2 = 12 b. x^2+y^2 = 36 c. x^2+(y + 1)^2 = 12 d. x^2+(y + 1)^2 = 36 15. what is the radius of the circle with the equation (x - 10)^2+(y + 3)^2 = 1? a. 10 b. 2 c. 1 d. 0.5

Explanation:

Step1: Recall circle - equation formula

The standard - form equation of a circle is \((x - h)^2+(y - k)^2=r^2\), where \((h,k)\) is the center of the circle and \(r\) is the radius.

Question 13:

The center of the circle is \((h = 3,k=-2)\) and \(r = 4\).
Substitute \(h = 3,k=-2,r = 4\) into the standard - form equation:
\((x - 3)^2+(y-(-2))^2=4^2\), which simplifies to \((x - 3)^2+(y + 2)^2=16\).

Question 14:

The center of the circle is \((h = 0,k=-1)\) and \(r = 6\).
Substitute \(h = 0,k=-1,r = 6\) into the standard - form equation:
\((x - 0)^2+(y-(-1))^2=6^2\), which simplifies to \(x^2+(y + 1)^2=36\).

Question 15:

The equation of the circle is \((x - 10)^2+(y + 8)^2=1\).
Comparing with the standard - form equation \((x - h)^2+(y - k)^2=r^2\), we have \(r^2 = 1\), so \(r = 1\).

Answer:

  1. C. \((x - 3)^2+(y + 2)^2=16\)
  2. D. \(x^2+(y + 1)^2=36\)
  3. C. 1