Question 924394
Assume that the point is (x,y).
"abscissa of a point is two-fifths of its ordinate"
{{{x=(2/5)y}}}
Then use the distance formula,
{{{(x-(-2))^2+(y-2)^2=5^2}}}
{{{(x+2)^2+(y-2)^2=25}}}
Substituting,
{{{((2/5)y+2)^2+(y-2)^2=25}}}
{{{(4/25)y^2+(8/5)y+4+y^2-4y+4=25}}}
{{{(29/25)y^2-(12/5)y+8=25}}}
{{{29y^2-60y+200=625}}}
{{{29y^2-60y-425=0}}}
{{{(y-5)(29y+85)=0}}}
Two solutions:
{{{y-5=0}}}
{{{y=5}}}
Then,
{{{x=(2/5)5=2}}}
(2,5)
.
.
{{{29y+85=0}}}
{{{29y=-85}}}
{{{y=-(85/29)}}}
Then,
{{{x=(2/5)(-(85/29))=-34/29}}}
(-34/29,-85/29)
.
.
.
{{{drawing(300,300,-8,8,-8,8,grid(1),circle(-2,2,0.2),circle(2,5,0.2),circle(-34/29,-85/29,0.2),circle(2,5,5),circle(-34/29,-85/29,5),blue(line(-2,2,2,5)),blue(line(-2,2,-34/29,-85/29,0.2)))}}}