A circle has center two, two and goes through the point six, three. Find the equation of the circle.
The equation of a circle is 𝑥 minus 𝑎 all squared plus 𝑦 minus 𝑏 all squared is equal to 𝑟 squared, where the center has coordinates or ordered pair 𝑎, 𝑏 and the radius of the circle is length 𝑟.
In our example, we know that the center has coordinates two, two. As the point six, three lies on the outside of the circle, we know that the radius is the distance between six, three and two, two. The distance between two coordinates or points can be calculated by square-rooting 𝑥 one minus 𝑥 two all squared plus 𝑦 one minus 𝑦 two all squared.
Substituting in our two coordinates in this case will help us calculate the radius. The radius is equal to the square root of six minus two all squared plus three minus two all squared. Six minus two is equal to four. And four squared is 16. Three minus two is equal to one. And one squared is equal to one. Therefore, the radius is the square root of 16 plus one. As 16 plus one is 17, our radius is root 17.
As we now know the center and the radius of the circle, we can substitute these values into the equation of the circle formula. This gives us 𝑥 minus two all squared plus 𝑦 minus two all squared is equal to root 17 squared. As root 17 squared is equal to 17, we can say that the equation of the circle with center two, two that passes through the point six, three is equal to 𝑥 minus two all squared plus 𝑦 minus two all squared equals 17.