Lable each coordinates, (0,0), (3,4), (6,8), and (6,4), point O, P, Q and R respectively, where O(0,0) is the origin of the coodinates, P(3,4) is the center of the circle with radius 5 units long, Q(6,8) is a point supposed to be on the circle, and R(6.4) is the intersection of 2 lines, y=4 and x=6.
In ΔPQR, PR⊥QR, PR=3 units, and QR=4 units, so PQ=5 units. (3-4-5 triangle)
Therefore, point Q(6,8) lies on the given circle P(3,4) with radius 5 units.* Q.E.D.
* OP is also 5 units long, so the circle passes thru O. The slope of OP and PQ is identical (=4/3) to each other, so points O,P and Q are colinear, and OQ is a diameter of the circle. The equation of the circle is: (x-3)²+(y-4)²=5².