Question 1066238
When ( h,k ) is the center of the circle,
{{{ ( x - h )^2 + ( y - k )^2 = r^2 }}}
{{{ x^2 - 2h*x + h^2 + y^2 - 2k*y + k^2 - r^2 = 0 }}}
{{{ x^2 + y^2 -2h*x - 2k*y + ( h^2 + k^2 - r^2 ) = 0 }}}
Comparing this to the given equation:
(1) {{{ 2x = 2h*x }}}
(2) {{{ 4y = 2k*y }}}
(3) {{{ -11 = h^2 + k^2 - r^2 }}}
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(1) {{{ 2 = 2h }}}
(1) {{{ h = 1 }}}
and
(2) {{{ 4 = 2k }}}
(2) {{{ k = 2 }}}
and
{{{ -11 = 1^2 + 2^2 - r^2 }}}
{{{ r^2 = 11 + 1 + 4 }}}
{{{ r^2 = 16 }}}
{{{ r = 4 }}}
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The center is at ( h, k ) = ( 1, 2 )
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The x-intercept is where {{{ y = 0 }}}
{{{ ( x - h )^2 + ( y - k )^2 = r^2 }}}
{{{ ( x - 1 )^2 + ( 0 - 2 )^2 = 4^2 }}}
{{{ ( x - 1 )^2 = 16 - 4 }}}
{{{ ( x - 1 )^2 = 12 }}}
{{{ x - 1 = sqrt(12) }}}
{{{ x = 1 + 2*sqrt(3) }}}
and
{{{ x = 1 - 2*sqrt(3) }}}
Those are x-intercepts
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The y-intercepts are where {{{ x = 0 }}}
{{{ ( x - h )^2 + ( y - k )^2 = r^2 }}}
{{{ ( 0 - 1 )^2 + ( y - 2 )^2 = 4^2 }}}
{{{ 1^2 + ( y - 2 )^2 = 16 }}}
{{{ ( y - 2 )^2 = 16 - 1 }}}
{{{ ( y - 2 )^2 = 15 }}}
{{{ y - 2 = sqrt(15) }}}
{{{ y = 2 + sqrt(15) }}}
and
{{{ y = 2 - sqrt(15) }}}
Those are y-intercepts
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Here's the plot:
{{{ graph( 400, 400, -10, 10, -10, 10, sqrt( 16 - ( x-1 )^2 ) + 2, -sqrt( 16 - ( x-1 )^2 ) + 2 ) }}}