The equation of a circle is (x-2)^2+(y+3)^2=4?

Question

I have to tell whether each point is on the circle, interior, or exterior of the circle. The points are (2, - 4)
II also need it explained step by step

Thanks

Answer 1

The Center is (2,-3)
Radius is 2

Here is how you do it. If the distance from (2,-3) to (2,-4) is larger than 2, then its outside the circle, it its smaller than 2, then its inside the cirle, if its exact 2, then its on the circle.

(2,-3) and (2,-4)

D = sqrt((2 - 2)^2 + (-4 - (-3))^2)
D = sqrt(0^2 + (-1)^2)
D = sqrt(0 + 1)
D = sqrt(1)
D = 1

Since 1 is smaller than 2

ANS : Interior

Answer 2

All you have to do is substitute the coordinates (2, -4) in the equation where x and y are. Put the x coordinate in place of the x and the same with y. Then if the equation totals 4, it's on the circle. If not, you have to graph it.
The equation's total of 4 is equal to the radius squared. So the radius is 2. Then the center of the circle is determined by the (x-2) and (y+3). X-2 means that the center is 2 right from the 0,0 axis, or the origin. Y+3 means that the center is 3 up from the origin. So the center of the circle is (-2, 3) with a radius of 2. Graph and see where the point lies.

Answer 3

Well, lets look at this.
the equation for a circle is x^2 + y^2 = z^2
it looks like they substituted x-2 for x, y +3 for y, and 4 for z^2
that would mean that z = 2, right?
Going further, z is the radius, so we have a circle with a radius of 2.

that x - 2, hmm. if we had used x -0, then the circle would have a center at x = 0. In this case, the x - 2 means that you have to subract 2 from x to get to zero, so the circle is shifted to center at x=2.

next, we see the y+3. The circle must be raised by 3 to get to the origin, so it is centered at y = -3.

The circle is thus centered at (2,-3). You want to look at the point (2,-4) The distance from the center to the point is 1 (in the y direction). The radius of the circle is 2.

Does that help a little?

Basically, remember the formula for a circle, and shift the center according to the substitution values for x and y squared. Get the radius from the z squared term.

Good luck.

Answer 4

The point (2, -4) is WITHIN the circle.

Your circle equation: (x-2)^2+(y+3)^2=4 is ALMOST in standard form:

[x-h]^2 + [y-k]^2 = r^2 Where (h,k) are the circle-center coordinates and r is the radius of the circle.

A little "triggering" with your equation yields this:

[x-2]^2+[y-(-3)])^2 =2^2 so your circle-center is (2, -3); your radius is 2.

Very nicely the point in question (2, - 4) is DIRECTLY BELOW THE CENTER (and ONLY 1 unit away, less than the 2 unit radius), so quite easily you determine (2, -4) is WITHIN the circle.
- - - - - - - - -

Were the point in question not horizontally or vertically aligned with the circle-center, then you'd use the distance formula (between center-point and question-point) to determine whether the question point were larger (outside-the-circle) or smaller (within-the-circle) than the radius. [Or equal the radius -- this is almost trivial -- which implies the question point is ON the circle.]

distance formula = sqrt [ (x2-x1)^2 + (y2-y1)^2]

Answer 5

The standard form equation of a circle is a way to express the definition of a circle on the coordinate plane.

* On the coordiante plane, the formula becomes (X-H)2 + (Y-K)2=r2
***h and k are the x and y coordines of the center of the circle
***(x-9)2 + (y-6)2=100 is a circle centered at (9,6) with a radius of 10

Answer 6

the brackets tell you where the centre is:

let the brackets = 0
x - 2 = 0 x must be 2 repeat for y

x^2 + y^2 = r^2

r = radius = 2 in this case (from the centre to 2 units out)

so, the point 2, - 4 is not in the circle (try drawing this)

Answer 7

(2,-4) is only one point. Remember the song and dance about coordinates> Just plug it in and find out. I think you will find that the result is NOT 4, so the point is NOT on the circle.

Answer 8

examine although if the equation is suitable: permit M(x, y) = x + y permit N(x, y) = x - y ?M(x, y)/?y = a million ?N(x, y)/?x = a million The equation is suitable. as a result, there exists a function, F(x, y) the partial spinoff with admire to x is M(x, y) and the partial spinoff with admire to y is N(x, y). we start up by integrating M(x, y) with admire to x: ?(x + y)dx = (a million/2)x² + yx + h(y) the place h(y) is a function that could disappear as quickly as we differentiate with admire to x. We differentiate this with admire to y: x + h'(y) This could equivalent N(x, y); x + h'(y) = x - y h'(y) = -y h(y) = (-a million/2)y² + C the answer is: F(x, y) = (a million/2)(x² - y²) + xy + C