find all the points having an x-coordinate of 2 whose distance from the point (-2,-1) is 5.?

Question

AH its so hard! help mee pleasee!!

Answer 1

Distance from point = 5, this is a circle.

(x + 2 )^2 + (y+1)^2 = 25

Plug in x = 2
(4)^2 + (y+1)^2 = 25
(y+1)^2 = 25 -16 = 9
y+1 = +3 or -3
y = +2 or -4

Points are ( 2, 2) and (2, -4)

Answer 2

The solution is to use the equation of a circle (X-X1)^2 + (Y-Y1)^2 = R^2. where X1=(-2), Y1=(-1) and X=2. Then solving for Y gives the solution:

(2-(-2))^2 + (Y-(-1))^2 = (5)^2
or
16 + (Y+1)^2 = 25

So Y=2 or -4

Thus the Solution is (2, 2) and (2, -4).

It's easier if you make a picture in your mind. You need to "see" the problem. Here's how I "see" this.

Locate the point (-2, -1) on a graph. Can you see it? Stick a thumb tack in that point. Now find a piece of string which is exactly 5 units long. Attach the string to the thumb tack at (-2, -1). If you pull the string tightly and pull it around the thumb tack for 360 degrees it will create a circle. Can you see the circle in your mind whose center is (-2, -1) and whose radius is 5 units?

Remember that a circle is really a collection of points that are a fixed distance away from the radius (in our case 5 units), from the center (in our case at (-2, -1)).

Now, the question you have to answer is "find all the points" where X=2. Simple. In your imaginary circle, when X=2, what are the possible values for Y? Answer, 2 and -4.

If you want to "see" this go to the link listed in the source below.

Answer 3

All of the points whose distance from the point (-2,-1) is 5 are given by the equation (x+2)^2 + (y+1)^2 = 25.
If x=2, 16 +(y+1)^2 =25 --> (y+1)^2= 9 so y+1= +/- 3.
Thus y = -1+/- 3= 2 or -4.

Thus the answer is (2,2) and (2,-4).

Answer 4

this is the locus of a circle with radius 5, center at (-2,-1)
(x + 2 )^2 + (y+1)^2 = 25