Question

If a circle has a center at the point (3, 4) and goes through the point (6, 8), what is the radius?

Answer 1

5

Step-by-step explanation:

We know that the standard circle equation is as follows:

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

where,

`r` = radius of the circle

Let P = center of the circle and Q = any point on the circle

`\left ( h,k \right ) = P = \left ( 3,4 \right )`

`\left ( x,y \right ) = Q = \left ( 6,8 \right )`

Solution is as follows:

`r = \sqrt{(x-h)^(2) + (y-k)^(2)} \\`

`r= \sqrt{(6-3)^(2) + (8-4)^(2)}\\`

`r= \sqrt{(3)^(2) + (4)^(2)}\\`

`r= √(9+16)}\\`

`r=√(\\25)\\`

`r=5`

Answer 2

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.