Ask Question
10 November, 16:19

What is the radius of a circle whose equation is x2+y2+8x-6y+21=0?

+1
Answers (2)
  1. 10 November, 16:29
    0
    Radius=2

    Center: (-4,3)

    Step-by-step explanation:

    The general equation of a circle is

    (x-xo) ² + (y-yo) ²=r², where (xo, yo) are the coordinates of the center, and r is the radius.

    Now: x²+y²+8x-6y+21=0

    x²+8x+y²-6y+21=0

    (x²+8x+16-16) + (y²-6y+9-9) + 21=0

    (x²+8x+16) - 16 + (y²-6y+9) - 9+21=0

    (x+4) ² + (y-3) ²-16-9+21=0

    (x+4) ² + (y-3) ²=16+9-21=25-21=4=2²

    So

    (x+4) ² + (y-3) ²=2².

    Thus, the radius is 2.

    The center is at (-4,3)
  2. 10 November, 20:03
    0
    the radius is 2
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “What is the radius of a circle whose equation is x2+y2+8x-6y+21=0? ...” in 📘 Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers