A circle in the [tex]xy[/tex]-plane has a diameter with endpoints [tex]\((2,4)\)[/tex] and [tex]\((2,14)\)[/tex]. An equation of this circle is [tex]\((x-2)^2+(y-9)^2=r^2\)[/tex], where [tex]r[/tex] is a positive constant. What is the value of [tex]r[/tex]?



Answer :

To solve for the value of [tex]\( r \)[/tex] in the given circle equation, we need to find the radius of the circle.

Let's start by identifying the coordinates of the endpoints of the diameter: [tex]\((2, 4)\)[/tex] and [tex]\((2, 14)\)[/tex].

1. Find the distance between the two endpoints of the diameter:

The distance formula between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] in the plane is:
[tex]\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Plugging in our points:
[tex]\[ d = \sqrt{(2 - 2)^2 + (14 - 4)^2} \][/tex]
Since the [tex]\( x \)[/tex]-coordinates are the same, the formula simplifies to:
[tex]\[ d = \sqrt{0 + (14 - 4)^2} = \sqrt{10^2} = 10 \][/tex]

Thus, the diameter of the circle is [tex]\( 10 \)[/tex].

2. Calculate the radius of the circle:

The radius [tex]\( r \)[/tex] is half of the diameter. So,
[tex]\[ r = \frac{\text{diameter}}{2} = \frac{10}{2} = 5 \][/tex]

Therefore, the radius [tex]\( r \)[/tex] of the circle is [tex]\( 5 \)[/tex].

So, the value of [tex]\( r \)[/tex] is [tex]\( \boxed{5} \)[/tex].