Question 1175478
given:

centre C ({{{4}}},{{{1}}}) -> {{{h=4}}}, {{{k=1}}}

distance between C and D is equal to the radius

{{{r^2=(1-4)^2+(6-1)^2}}}..eq.1
{{{r^2=9+25}}}
{{{r^2=34}}}

formula of the circle is:

{{{(x-h)^2+(y-k)^2=r^2}}}
{{{(x-4)^2+(y-1)^2=34}}}..........eq.1

the endpoint of the radius of a circle is D ({{{1}}},{{{6}}})

DE is the diameter of the circle, so find equation of the line passing through the endpoint of the radius D ({{{1}}},{{{6}}}) and centre C ({{{4}}},{{{1}}}) 

{{{y=mx+b}}}

use points to find a slope:

{{{m=(1-6)/(4-1)=-5/3}}}
so far equation is

{{{y=-(5/3)x+b }}}.....use one point to calculate {{{b}}}

{{{1=-(5/3)4+b}}}

{{{1=-20/3+b}}}

{{{b=1+20/3}}}

{{{b=23/3}}}
and equation of the line is

{{{y=-(5/3)x+23/3}}}..........eq.2

the endpoint E ({{{x}}},{{{y}}}) will be intersection point of the circle and line

so, to find it solve this system

{{{(x-4)^2+(y-1)^2=34}}}..........eq.1
{{{y=-(5/3)x+23/3}}}..........eq.2
-------------------------------------
substitute {{{y}}} from eq.2 in eq.1 and solve for {{{x}}}

{{{(x-4)^2+(-(5/3)x+23/3-1)^2=34}}}
{{{(x-4)^2+(-(5/3)x+23/3-1)^2=34}}}
{{{34/9 (x - 4)^2 = 34}}}
 {{{(x - 4)^2 = 34*9/34}}}
 {{{(x - 4)^2 = 9}}}
{{{(x - 4)=3}}} or {{{(x - 4)=-3}}}

solutions:

{{{x =7 }}}or {{{x =1}}}

go to
{{{y=-(5/3)x+23/3}}}..........eq.2, substitute {{{x}}}

if {{{x=7}}}
{{{y=-(5/3)7+23/3}}}
{{{y=-4}}}


if {{{x=1}}}

{{{y=-(5/3)1+23/3}}}
{{{y=6}}}

intersection points are: ({{{7}}},{{{-4}}}) and ({{{1}}},{{{6}}})

since given that D ({{{1}}},{{{6}}}), then E ({{{7}}},{{{-4}}})


{{{drawing( 600, 600, -10, 10, -10, 10,
circle(7,-4,.12),circle(1,6,.12),circle(4,1,.12),
locate(7.5,-4,E(7,-4)),locate(1.5,6,D(1,6)),locate(4.5,1,C(4,1)),
 graph( 600, 600, -10, 10, -10, 10, sqrt(34-(x-4)^2)+1, -sqrt(34-(x-4)^2)+1,-(5/3)x+23/3) )}}}