Does anyone know how to find the equation of a circle with the center (-1,9) and a radius of √3.?

Question

Does anyone know how to find the equation of a circle with the center (-1,9) and a radius of √3.

Answer 1

using (x-a)^2+(y-b)^2 = r^2
we get
(x+1)^2+(y-9)^2 = 3
or x^2+2x+1 + y^2 -18y + 81 = 3

or x^2+2x+ y^2 -18y + 79 = 0

Answer 2

EQUATION OF A CIRCLE
The equation of a circle comes in two forms:

1) The standard form: (x - h)2 + (y-k)2 = r
2) The general form : x2 + y2 + Dx + Ey + F = 0, where D, E, F are constants.

If the equation of a circle is in the standard form, we can easily identify the center of the circle, (h, k), and the radius, r . Note: The radius, r, is always positive.

Answer 3

the general equation of a circle is:
(x-a)^2 + (y-b)^2 = r^2, where (a,b) is the center and r is radius

so in this case, the equation would be
(x+1)^2 + (y-9)^2 = 3.

Hope that helps.

Answer 4

x^2 + y^2 = r^2

If the center is (-1,9), you just offset inside the exponent.
Thus,

(x+1)^2 + (y-9)^2 = 3

Answer 5

Circle Formula
(x - h)^2 + (y - k)^2 = r^2

(x - (-1))^2 + (y - 9)^2 = (sqrt(3))^2
(x + 1)^2 + (y - 9)^2 = 3

Answer 6

Sure. It is (x+1)^2 + (y-9)^2 = 3. Study this example, and see how it works; then no problem of this sort will ever bug you again.

Answer 7

Take Log of the two factors , remembering that Loga^n = nLog(a) so we get xLog(5/4) = Log(one hundred twenty five/sixty 4); divide the two factors via making use of Log(5/4) to get x = Log(one hundred twenty five/sixty 4) Log(sixty 4). indoors the numerator, one hundred twenty five = 5^3, sixty 4 = 4^3 so the numerator will become 3Log(5/4) so dividing via making use of Log(5/4) we get x = 3