Answer:
Point E could be any point on the circumference of the circle [tex](x+2)^2+(y-6)^2=100[/tex]
Step-by-step explanation:
Let
(x,y) ----> the coordinates of point E
we know that
the formula to calculate the distance between two points is equal to
[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]
we have
[tex]points\ D(-2,6),E(x,y)[/tex]
[tex]d_E_D=10\ units[/tex]
substitute in the formula
[tex]10=\sqrt{(y-6)^{2}+(x+2)^{2}}[/tex]
squared both sides
[tex]100=(y-6)^2+(x+2)^2[/tex]
Rewrite
[tex](x+2)^2+(y-6)^2=100[/tex]
This is the equation of a circle with center at (-2,6) and radius equal to 10 units
therefore
Point E could be any point on the circumference of the circle [tex](x+2)^2+(y-6)^2=100[/tex]
