Question 758531
Find the coordinates of a point on a circle at the opposite end of the diameter from 
point A ({{{-7}}},{{{-1}}})? 
The circle's formula is {{{(x+4)^2+(y-3)^2=25}}}

from the circle's formula we can see that radius {{{r=5}}}; so, diameter is {{{d=10}}}, the center is at ({{{-4}}},{{{3}}})

now we can find equation of the line passing through points A ({{{-7}}},{{{-1}}}) and ({{{-4}}},{{{3}}}):

*[invoke change_this_name10094 -7, -1, -4, 3]

if you go three units to the right and four units up from the point ({{{-7}}},{{{-1}}}), you will get to the point ({{{-7+3}}},{{{-1+4}}})=({{{-4}}},{{{3}}}) and distance will be {{{5}}} which is radius

then go three units to the right and four units up from the center ({{{-4+3}}},{{{3+7}}})=({{{-1}}},{{{7}}}) and distance will be {{{10}}}

or,

If you know the center of the circle, then it is the midpoint of the diameter. You can use the midpoint formula to solve for the {{{x}}} and {{{y}}} values of the other endpoint of the diameter.


{{{x[mid] =(x1 + x2)/2 }}}
{{{y[mid] =( y1 + y2)/2}}}
-------------------------
{{{-4 =(-7 + x2)/2 }}}
{{{3 =(-1 + y2)/2}}}
---------------------
{{{-8 =(-7 + x2) }}}=> {{{-8+7=x2 }}}=> {{{-1=x2 }}}

{{{6 =(-1 + y2)}}}=>{{{6 +1=y2)}}}=>{{{7=y2)}}}

so, the point is ({{{-1}}},{{{7}}})



{{{ drawing( 600, 600, -10, 10, -10, 10,blue(line(-7,-1,-4,-1)),blue(line(-4,3,-4,-1)),blue(line(-4,3,-1,3)),blue(line(-1,3,-1,7)),circle(-4,3,0.1),circle(-7,-1,0.1),circle(-1,7,0.1),graph( 600, 600, -10, 10, -10, 10,(1.33333333333333)x + 8.33333333333333, sqrt(-(x+4)^2+25)+3,-sqrt(-(x+4)^2+25)+3)) }}}