MHB Find the Radius and Center of a Circle

[squexy]

Can someone show me how to resolve this question?

A particular circle in the standard (x,y) coordinate
plane has an equation of (x − 5)2 + y2 = 38. What are
the radius of the circle, in coordinate units, and the
coordinates of the center of the circle?
radius center
F. √38 ( 5,0)
G. 19 ( 5,0)
H. 38 ( 5,0)
J. √38 (−5,0)
K. 19 (−5,0)The answer is F

 

[I like Serena]

Can someone show me how to resolve this question?

A particular circle in the standard (x,y) coordinate
plane has an equation of $(x − 5)^2 + y^2 = 38$. What are
the radius of the circle, in coordinate units, and the
coordinates of the center of the circle?
radius center
F. √38 ( 5,0)
G. 19 ( 5,0)
H. 38 ( 5,0)
J. √38 (−5,0)
K. 19 (−5,0)The answer is F

Hi squexy! Welcome to MHB! :)

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

Your equation can be written as:
$$(x − 5)^2 + (y-0)^2 = (\sqrt{38})^2$$

This shows that the radius is $\sqrt{38}$ and the center is $(5,0)$.

 

[squexy]

Hi squexy! Welcome to MHB! :)

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

Your equation can be written as:
$$(x − 5)^2 + (y-0)^2 = (\sqrt{38})^2$$

This shows that the radius is $\sqrt{38}$ and the center is $(5,0)$.

Thank you for the answer.

Why is 38 in square root?
(x−5)2+(y−0)2= (√38 )²

Since the equation given is (x−5)2+y2=38. shouldn't the equation transferred to the circle formula be written as:

(x−5)2+(y−0)2= (38 )²

 

[I like Serena]

Thank you for the answer.

Why is 38 in square root?
(x−5)2+(y−0)2= (√38 )²

Since the equation given is (x−5)2+y2=38. shouldn't the equation transferred to the circle formula be written as:

(x−5)2+(y−0)2= (38 )²

Well, $(38)^2 = 38 \cdot 38 = 1444$. This does not match the $38$ in the original equation.

What we need is a number for $r$ that when squared gives us $38$. This is what the square root is. It means that $(\sqrt 38)^2 = \sqrt 38 \cdot \sqrt 38 = 38$, which is what is required.

 

[Deveno]

Look again at post #2.

The form we should have:

$(x - a)^2 + (y - b)^2 = r^2$

What we do have:

$(x - 5)^2 + y^2 = 38$.

It's clear from this that the "$x$" part is already what we need, so $a = 5$.

The "$y$" part is a bit trickier, we solve:

$(y - b)^2 = y^2$

$y^2 - 2by + b^2 = y^2$.

$b(b - 2y) = 0$. <---this has to be true for ALL $y$, and not just when $y = \dfrac{b}{2}$ ($b$ is a constant).

We conclude that $b = 0$.

The only thing left to do is solve:

$r^2 = 38$.

This happens when $r = \pm \sqrt{38}$.

Since a radius is always positive, we take the positive square root for $r$.

The number on the RHS of the standard circle equation is the radius SQUARED, not the radius. For example, a circle of radius 3, centered at the origin has equation:

$x^2 + y^2 = 9$ <---note we have 9, not 3.

$(9,0)$ does not lie on this circle, since:

$9^2 + 0^2 = 81 \neq 9$.

$(3,0)$ DOES lie on this circle, since:

$3^2 + 0^2 = 9$.

 

[squexy]

Well, $(38)^2 = 38 \cdot 38 = 1444$. This does not match the $38$ in the original equation.

What we need is a number for $r$ that when squared gives us $38$. This is what the square root is. It means that $(\sqrt 38)^2 = \sqrt 38 \cdot \sqrt 38 = 38$, which is what is required.

Thanks ^.^