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# Given eqn of circle, find the range?

### 3 Answers

- 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?

Done!