Question 1078546

 Find the equation of the circle that passes through ({{{-1}}},{{{1}}}) and with center at the point of intersection of {{{x + 3y + 7 = 0}}} and {{{3x - 2y - 12 = 0}}}.


equation of the circle: 

{{{(x-h)^2+(y-k)^2=r^2 where {{{h}}} and {{{k}}} are {{{x}}} and {{{y}}} coordinates of center, and {{{r}}} is radius

use {{{x + 3y + 7 = 0}}} and {{{3x - 2y - 12 = 0}}} to find  {{{x}}} and {{{y}}} coordinates of center

{{{x + 3y + 7 = 0}}}...........solve for {{{x}}}
{{{x =-3y-7}}}............substitute in equation

{{{3(-3y-7) - 2y - 12 = 0}}}

{{{-9y -21- 2y - 12 = 0}}}

{{{-11y -33 = 0}}}
{{{-11y = 33}}}
{{{y = 33/-11}}}
{{{y = -3}}}

find {{{x}}

{{{x =-3(-3)-7}}}
{{{x =9-7}}}
{{{x =2}}}

so, {{{x=h}}} and {{{y=k}}} coordinates of center are: {{{2}}} and {{{-3}}}

and equation of your circle (so far) is: 
{{{(x-2)^2+(y-(-3))^2=r^2
{{{(x-2)^2+(y+3)^2=r^2............now we need a radius {{{r}}}

use given point  ({{{-1}}},{{{1}}}) and substitute its {{{x}}} and {{{y}}} coordinates

{{{(-1-2)^2+(1+3)^2=r^2}}}
{{{(-3)^2+(4)^2=r^2}}}
{{{9+16=r^2}}}
{{{25=r^2}}}

{{{(x-2)^2+(y+3)^2=25 }}} ->equation of your circle



{{{drawing( 600, 600, -15, 15, -15, 15,
circle(2,-3,.2), locate(2,-3,C(2,-3)),
circle(-1,1,.2), locate(-1,1,p(-1,1)),
 graph( 600, 600, -15, 15, -15, 15, -x/3-7/3, 2x/2-6,sqrt(-(x-2)^2+25)-3,-sqrt(-(x-2)^2+25)-3)) }}}