We know that the standard form of a circle is given as:
(x – h)^2 + (y – k)^2 = r^2
where the variables are,
h and k are the center points of the circle
x and y are points on the circle
r is the radius
In this case, we are given that:
(h, k) = (3, 4)
(x, y) = (6, 8)
Therefore substituting these values into the equation and calculating for radius, r:
(6 – 3)^2 + (8 – 4)^2 = r^2
r^2 = 3^2 + 4^2
r^2 = 9 + 16
r^2 = 25
r = 5
Therefore the radius of the circle is 5 units.
Answer:
5
Explanation:
We know that the standard circle equation is as follows:
[tex]r^{2}=(x-h)^{2} + (y-k)^{2}[/tex]
where,
[tex]r[/tex] = radius of the circle
Let P = center of the circle and Q = any point on the circle
[tex]\left ( h,k \right ) = P = \left ( 3,4 \right )[/tex]
[tex]\left ( x,y \right ) = Q = \left ( 6,8 \right )[/tex]
Solution is as follows:
[tex]r = \sqrt{(x-h)^{2} + (y-k)^{2}} \\[/tex]
[tex]r= \sqrt{(6-3)^{2} + (8-4)^{2}}\\[/tex]
[tex]r= \sqrt{(3)^{2} + (4)^{2}}\\[/tex]
[tex]r= \sqrt{9+16}}\\[/tex]
[tex]r=\sqrt{\\25}\\[/tex]
[tex]r=5[/tex]
