Find Equation of Point P Moving at Distance 4 from (-2,3)

[ramz]

Hello everyone.. I need your help to solve this problem.

A point P(x,y) moves in such a way that its distance from (-2, 3) is 4. Find the equation.

Thanks

 

[MarkFL]

Do we have any kind of formula for the distance between two points in a plane?

 

[ramz]

d² = (x₂-x₁)²+(y₂-y₁)²

 

[MarkFL]

d² = (x₂-x₁)²+(y₂-y₁)²

Yes, now can you identify $d$, and the two points, one fixed and the other arbitrary?

 

[ramz]

Yes, now can you identify $d$, and the two points, one fixed and the other arbitrary?

Maybe, I don't think so. .

 

[MarkFL]

Maybe, I don't think so. .

Well, let's start with the distance $d$ we want to exist between the two points...what should that be? What does the problem say it should be?

 

[ramz]

Well, let's start with the distance $d$ we want to exist between the two points...what should that be? What does the problem say it should be?

I don't know. Honestly, I'm confused with the problem.

 

[MarkFL]

I don't know. Honestly, I'm confused with the problem.

Let's look at the problem statement:

A point P(x,y) moves in such a way that its distance from (-2, 3) is 4. Find the equation.

I have highlighted the information regarding what the distance is to be...so plug that in for $d$, and what do we have now?

 

[ramz]

Let's look at the problem statement:

A point P(x,y) moves in such a way that its distance from (-2, 3) is 4. Find the equation.

I have highlighted the information regarding what the distance is to be...so plug that in for $d$, and what do we have now?

Is this the next move?
4²=(x+2)²+(y-3)²

 

[MarkFL]

Is this the next move?
4²=(x+2)²+(y-3)²

Yes, that's it!

The distance $d$ is 4, and you have let:

\(\displaystyle \left(x_2,y_2\right)=(x,y)\)

\(\displaystyle \left(x_1,y_1\right)=(-2,3)\)

You could of course reverse the two points. Do you recognize the equation you have derived? What does it represent? What plane curve is defined as the set of all points equidistant from some central point?

 

[ramz]

Yes, that's it!

The distance $d$ is 4, and you have let:

\(\displaystyle \left(x_2,y_2\right)=(x,y)\)

\(\displaystyle \left(x_1,y_1\right)=(-2,3)\)

You could of course reverse the two points. Do you recognize the equation you have derived? What does it represent? What plane curve is defined as the set of all points equidistant from some central point?

x²+y²+4x+6y-3 = 0
Equation of a circle.
So, what could be the points?

 

[MarkFL]

I would leave the equation in the standard form:

\(\displaystyle (x-h)^2+(y-k)^2=r^2\)

where the center of of circle is $(h,k)$ and the radius is $r$. All of the points $(x,y)$ on the circle will satisfy the given equation. Let's examine our equation:

\(\displaystyle (x+2)^2+(y-3)^2=4^2\)

We know that the point 4 units directly above the center is on the circle...and this point is $(x,y)=(-2,7)$ so we should be able to plug this into the equation and get an identity (something that is true). So, let's try it:

\(\displaystyle (-2+2)^2+(7-3)^2=4^2\)

\(\displaystyle 0^2+4^2=4^2\)

\(\displaystyle 4^2=4^2\)

This is true. :D

 

[suluclac]

The distance from (-2, 3) is 4 is the same thing as saying a circle with radius with its center at (-2, 3).
Although not necessarily true, it states clearly that a point P(x, y) moves in such a way.
A nice circle geometry problem, indeed!