Determine the equation of the circle: With x intercepts (2 ; 0) and (-4 ; 0) and a radius of 5

[john458776]

I'm really confused how I can approach this problem. Give me hints then I'll solve it and send the solution here. Please guide me and teach me.

 

[Harry_the_cat]

Have you drawn a diagram?

 

[Dr.Peterson]

I'm really confused how I can approach this problem. Give me hints then I'll solve it and send the solution here. Please guide me and teach me.

The center will be 5 units from each intercept. You can solve geometrically or algebraically; take your pick. If you draw the diagram right, you may be able to solve it mentally.

Of course, there are two such circles.

 

[john458776]

The center will be 5 units from each intercept. You can solve geometrically or algebraically; take your pick. If you draw the diagram right, you may be able to solve it mentally.

Of course, there are two such circles.

Okay, I tried finding x using the distance formular.

 

[Dr.Peterson]

Okay, I tried finding x using the distance formular.

I suppose you meant \(\displaystyle r^2=(x_1-x_2)^2+(y_1-y_2)^2\).

One algebraic approach would be to call the center (x, y) and use the distance formula to write two equations expressing the fact that this point is 5 units from each intercept. Have you done that?

The geometric approach, starting with a picture, is a lot simpler, especially if you know about the 3-4-5 triangle.

 

[skeeter]

Okay, I tried finding x using the distance formular.

The x-value of the circle center is midway between the x-values of the two x-intercepts ...

 

[john458776]

Okay, I tried finding x using the distance formular.
r² [MATH]=[/MATH] (x₂[MATH]-[/MATH]x₁)²[MATH]+[/MATH](y₂[MATH]-[/MATH]y₁)²
5² [MATH]=[/MATH] (x₂[MATH]-[/MATH]2)²[MATH]+[/MATH](y₂[MATH]-[/MATH]0)²
[MATH]21=[/MATH]x²[MATH]-4x+[/MATH]y² equation (1)

Then
r² [MATH]=[/MATH] (x₂[MATH]-[/MATH]x₁)²[MATH]+[/MATH](y₂[MATH]-[/MATH]y₁)²
5² [MATH]=[/MATH] (x₂[MATH]-[/MATH]2)²[MATH]+[/MATH](y₂[MATH]-[/MATH]0)²
[MATH]9=[/MATH]x²[MATH]+8x+[/MATH]y²
y²[MATH]=9-[/MATH]x²[MATH]-8x[/MATH] equation (2)
Substituting equation (2) into (1) I get [MATH]x=-1[/MATH]

 

[john458776]

The x-value of the circle center is midway between the x-values of the two x-intercepts ...

Okay I get it now.

Let's use the midpoint formular to find x-intercept
x_m[MATH] = ([/MATH]x₂[MATH]+[/MATH]x₁)/2
x_m[MATH]=[/MATH][MATH](2+(-4))[/MATH]/2
x_m[MATH]=-1[/MATH]

 

[pka]

Let's use the midpoint formular to find x-intercept
x_m[MATH] = ([/MATH]x₂[MATH]+[/MATH]x₁)/2
x_m[MATH]=[/MATH][MATH](2+(-4))[/MATH]/2
x_m[MATH]=-1[/MATH]

So the center of the circle is \((-1,k)\). Find \(k\) so that the center is \(5\) units to each of the intercepts.

 

[john458776]

So the center of the circle is \((-1,k)\). Find \(k\) so that the center is \(5\) units to each of the intercepts.

[MATH](x-a)[/MATH]²[MATH]+[/MATH][MATH](x-k)[/MATH]²[MATH]=[/MATH]5²

[MATH](2 ; 0)[/MATH]
[MATH](2+1)[/MATH]²[MATH]+[/MATH][MATH](0-k)[/MATH]²[MATH]=25[/MATH][MATH]9+[/MATH][MATH]k[/MATH]²[MATH]=25[/MATH][MATH]k=4[/MATH] or [MATH]k=-4[/MATH]
This is correct right ?
Which one is the y-intercept of center between [MATH]-4[/MATH] or [MATH]4[/MATH]

 

[pka]

[MATH]k=4[/MATH] or [MATH]k=-4[/MATH]This is correct right ?
Which one is the y-intercept of center between [MATH]-4[/MATH] or [MATH]4[/MATH]

Both are correct. As Prof. Peterson told you there are two circles that have intercepts \((-4,0)~\&~(2,0)\) with radius \(5\).

 

[Dr.Peterson]

[MATH](x-a)[/MATH]²[MATH]+[/MATH][MATH](x-k)[/MATH]²[MATH]=[/MATH]5²

[MATH](2 ; 0)[/MATH]
[MATH](2+1)[/MATH]²[MATH]+[/MATH][MATH](0-k)[/MATH]²[MATH]=25[/MATH][MATH]9+[/MATH][MATH]k[/MATH]²[MATH]=25[/MATH][MATH]k=4[/MATH] or [MATH]k=-4[/MATH]
This is correct right ?
Which one is the y-intercept of center between [MATH]-4[/MATH] or [MATH]4[/MATH]

Have you tried a drawing yet?



Now you have to write the equation of each circle. And, of course, you didn't mean y-intercept, just center.