Answer:
The equation of the circle can be written as:
Step-by-step explanation:
The general equation of a circle with center
and radius
is:
![\left(x-h\right)^2+\left(y-k\right)^2=r^2](https://tex.z-dn.net/?f=%5Cleft%28x-h%5Cright%29%5E2%2B%5Cleft%28y-k%5Cright%29%5E2%3Dr%5E2)
In our example, we know
, as we just have to make sure we need determine
.
![\left(x-4\right)^2+\left(y-4\right)^2=r^2\:\:](https://tex.z-dn.net/?f=%5Cleft%28x-4%5Cright%29%5E2%2B%5Cleft%28y-4%5Cright%29%5E2%3Dr%5E2%5C%3A%5C%3A)
As the circle passes through (10, 14), that pair of values for x and y must satisfy the equation. So we have:
![\left(10-4\right)^2+\left(14-4\right)^2=r^2](https://tex.z-dn.net/?f=%5Cleft%2810-4%5Cright%29%5E2%2B%5Cleft%2814-4%5Cright%29%5E2%3Dr%5E2)
![\mathrm{Switch\:sides}](https://tex.z-dn.net/?f=%5Cmathrm%7BSwitch%5C%3Asides%7D)
![r^2=\left(10-4\right)^2+\left(14-4\right)^2](https://tex.z-dn.net/?f=r%5E2%3D%5Cleft%2810-4%5Cright%29%5E2%2B%5Cleft%2814-4%5Cright%29%5E2)
![r^2=6^2+10^2](https://tex.z-dn.net/?f=r%5E2%3D6%5E2%2B10%5E2)
![r^2=36+100](https://tex.z-dn.net/?f=r%5E2%3D36%2B100)
![r^2=136](https://tex.z-dn.net/?f=r%5E2%3D136)
Thus the equation of the circle can be written as:
![\left(x-4\right)^2+\left(y-4\right)^2=136](https://tex.z-dn.net/?f=%5Cleft%28x-4%5Cright%29%5E2%2B%5Cleft%28y-4%5Cright%29%5E2%3D136)
Answer:
Subtract y from both sides of the equation.
x=8−y
2x+2y=16
Replace all occurrences of x with 8−y in each equation.
16=16
x=8−y
Remove any equations from the system that are always true.
x=8−y
Step-by-step explanation:
To find the answer, use pythagorean theorum (a^2 + b^2 = c^2)
We'll square a and c, and then subtract a from c.
9 + b^2 = 361
361 - 9 = 352
The square root of 352 is 18.7
18.7 should be the correct answer.