Algebra

Let's note something important before we begin. 

All the points 5 units away from (2, 7) forms a circle with radius 5 and center of (2, 7). 

A circle with that equation would look something like

\((x-2)^2+(y-7)^2 = 5^2\)

 

Combining that with the second part of the problem, we have a system with 

\((x-2)^2+(y-7)^2 = 25\\y=5x-28\)

 

Now, we already have a value for y, and we subsitute that into the first equation to isolate x. 

We find that

\((x-2)^2+(5x-35)^2 = 25\)

 

Expanding and bringing all terms to one side of the equation, we get a quadratic. We find that

\(26x^2-354x+1204=0\)

This gives us 2 x values that will satisfy thsi equation. We get that

\(x=7,\:x=\frac{86}{13}\)

 

Now, we plug these two values of x back into the equation to find y, so that we can complete our problem. 

We find that \(y = 7, y= \frac{66}{13}\)

 

So our final two abswers that satisfy all these conditions given above in the problem are

\((7, 7), (86/13, 66/13)\)

 

Thanks! :)