Yahoo Answers is shutting down on 4 May 2021 (Eastern Time) and, as of 20 April 2021 (Eastern Time), the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Given eqn of circle, find the range?
- PopeLv 72 months agoFavourite answer
Your circle equation is in this general form:
x² + y² + Dx + Ey + F = 0
center: (-D/2, -E,2)
radius = √(D² + E² - 4F) / 2
For this to be a circle, the radius must be real and positive.
D² + E² - 4F > 0
In your case:
(-6)² + (8)² - 4c > 0
100 - 4c > 0
4c < 100
c < 25
But that would have to be coupled with the given condition, c ≥ 0. Together, that places c on this interval:
0 ≤ c < 25
- az_lenderLv 72 months ago
(x-3)^2 = x^2 - 6x + 9.
(y+4)^2 = y^2 + 8x + 16.
So (x-3)^2 + (y+4)^2
= x^2 + y^2 - 6x + 8x + 25.
Therefore, if c = 25, the whole thing collapses to a single point (3,-4). If c < 25, you can move (c - 25) to the right-hand side, where it will represent r^2 for the circle. It's OK if C is negative, you'll get a bigger circle.
If c > 25 the equation cannot be satisfied.
- rotchmLv 72 months ago
Complete the square to get
(x-3)² + (y+4)² - 25 + C = 0. Convince yourself of this. Then rewrite as
(x-3)² + (y+4)² = 25 - C.
For this to be a circle, the RHS must be > 0.
25 - C > 0. Conclusion?