+2446 \((x-h)^2+(y-k)^2=r^2\) is the center-radius form for a circle such that \((h,k)\) is the coordinates of the center of the circle on the coordinate plane and \(r\) is the radius.
1) Find the equation of the circle centered at (1,-4) with radius 3
Since \((h,k)\) of \((x-h)^2+(y-k)^2=r^2\) represents the coordinates of the center and \((1,-4)\) is already known to be the center, then we know that \(h=1\text{ and }k=-4\) . \(r\) is the radius, so r=3.
Substitute these values into the center-radius form of a circle to generate the equation of this particular circle.
\(h=1,k=-4,r=3\\ (x-h)^2+(y-k)^2=r^2\\ (x-1)^2+(y-(-4))^2=3^2\\ (x-1)^2+(y+4)^2=9\)
2) I encourage you to use the information I gave you above to try and work out the second problem.
3) Find the distance between the points \((0,8)\text{ and }(2,-5)\).
We can use the distance formula, \(d=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}\) . Substitute the points that we want to find the distance from.
| \(d=\sqrt{(8-(-5))^2+(0-2)^2}\\ d=\sqrt{13^2+(-2)^2}\\ d=\sqrt{169+4}\\ d=\sqrt{173}\) | Substitute in the known coordinates and simplify completely. 173 has no perfect square factors, so the radical cannot be simplified any further. |
pls help 
