Question 329133
Use the distance formula. 
The distance between two points only yields 1 number.
{{{D^2=(x1-x2)^2+(y1-y2)^2}}}
{{{D^2=(-3-(-14))^2+(22-35)^2}}}
{{{D^2=(-3+14)^2+(-13)^2}}}
{{{D^2=(11)^2+(-13)^2}}}
{{{D^2=121+169}}}
{{{D^2=290}}}
{{{D=sqrt(290)}}}
The distance between ({{{-3}}}, {{{22}}}) and ({{{-14}}}, {{{35}}}) is {{{sqrt(290)}}}.
Follow the same procedure to get the distance for the remaining problems.
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The distance from the center of the circle to any point on the circle is equal to the radius of the circle. 
Again use the distance formula, but the distance you're calculating now is equal to the radius of the circle.
Again, the radius is only one number.
{{{R^2=(xc-x)^2+(yc-y)^2}}}
where ({{{xc}}},{{{yc}}}) is the center of the circle and ({{{x}}},{{{y}}}) is the point on the circle.
{{{R^2=(0-(-8))^2+(0-0)^2}}}
{{{R^2=8^2}}}
{{{R=8}}}
The circle centered at ({{{0}}},{{{0}}}) with point ({{{-8}}},{{{0}}}) on the circle has a radius of {{{8}}}.