So x minus the x coordinate of the center. Because both h and v are zero, they can disappear and you can simplify the standard circle equation to look like. So the x coordinate of the center must be negative five.
Gradient of QR. Features of a circle from its standard equation. And the way it's drawn right now, we could drag this out like this, but this the way it's drawn, the radius is indeed equal to two. The system of circles passes through P, Q.
Mid-point of PQ. Writing standard equation of a circle.
Notice we can construct a nice little right triangle here. Find the equation of the circle passing through the points P 2,1 , Q 0,5 , R -1,2.
Find the equations of perpendicular bisectors using gradient-point form. Equation of the circle can be found by substituting the coordinates of the points into the standard equation for a circle yielding three equations. Figure 3: Hopefully that makes sense.
Use Opposite angles of cyclic quad. Unfortunately, while it is much easier to graph circles at the origin, very few are as straightforward and simple as those.
Solve a and b. So you want to find all the x's and y's that are two away from it. Gradient of PQ. System of circles.
Mid-point of QR. This is not just, I don't want you to just memorize this formula. Eliminating F and get a simultaneous equation in D and E. Remember, if you have some center, in this case is the point negative five comma five, so negative five comma five, and you want to find all of the x's and y's that are two away from it.