Question 1140775
(1) A cirlce is defined by the equation {{{x^2-6x+y^2-2y=6}}}. 

(a) Rewrite the equation of the circle in the form:

{{{(x-a)^2+(y-b)^2=r^2 }}}

{{{(x^2-6x)+(y^2-2y)=6}}}............complete square

{{{(x^2-6x+b^2)-b^2+(y^2-2y+b^2)-b^2=6}}}.......{{{b}}} for {{{x}}} part is {{{6/2=3}}} and {{{b}}} for {{{y}}} part is {{{2/2=1}}}

{{{(x^2-6x+3^2)-3^2+(y^2-2y+1^2)-1^2=6}}}

{{{(x-3)^2-9+(y-1)^2-1=6}}}

{{{(x-3)^2+(y-1)^2-10=6}}}

{{{(x-3)^2+(y-1)^2=6+10}}}

{{{(x-3)^2+(y-1)^2=16}}}

{{{(x-3)^2+(y-1)^2=4^2}}}


(b) Write down the coordinates of M,the centre of the circle. 

since {{{a=3}}} and {{{b=1}}},

the coordinates of M are: ({{{a}}},{{{b}}})=({{{3}}},{{{1}}})


(c) A tangent PQ is drawn from the point P({{{6}}};{{{-2}}}) to the circle, with {{{Q}}} the point of contact.Calculate the length of the tangent {{{PQ}}}. 

if {{{PQ}}} is a tangent, it is perpendicular to the radius{{{ PM}}}

points {{{P}}},{{{Q}}}, and {{{M}}} form right triangle where {{{QM}}} is hypotenuse, {{{PM}}} and {{{PQ}}} are legs

we can find the length of {{{PM }}}using points P({{{6}}};{{{-2}}})  and M({{{3}}},{{{1}}})

{{{PM=sqrt((3-6)^2+(1-(-2))^2)}}}

{{{PM=sqrt((-3)^2+(1+2)^2)}}}

{{{PM=sqrt((-3)^2+(3)^2)}}}

{{{PM=sqrt(9+9)}}}

{{{PM=sqrt(2*9)}}}

{{{PM=3sqrt(2)}}}->one leg

and 

{{{PQ}}}, the other leg will be: 

P({{{6}}};{{{-2}}}) and Q({{{x}}},{{{y}}})

{{{PQ=sqrt((x-6)^2+(y+2)^2) }}}

then we can find hypotenuse {{{QM}}}:

{{{M}}}({{{3}}},{{{1}}}) and {{{Q}}}({{{x}}},{{{y}}})

{{{QM=sqrt((x-3)^2+(y-1)^2)}}} 


using Pythagorean theorem:

{{{(sqrt((x-3)^2+(y-1)^2) )^2=(sqrt((x-6)^2+(y+2)^2) )^2+(3sqrt(2))^2}}}

{{{(x-3)^2+(y-1)^2=(x-6)^2+(y+2)^2+18}}}

{{{x^2-6x+9+y^2-2y+1=x^2-12x+36+y^2+4y+4+18}}}

{{{-6x-2y+10=-12x+4y+58}}}

{{{-6x+12x=2y+4y+58-10}}}

{{{6x=6y+48}}}

{{{x=y+8}}}.............=>so point {{{Q}}}({{{y+8}}},{{{y}}})



then substitute in Pythagorean theorem:

{{{(sqrt((y+8-3)^2+(y-1)^2) )^2=(sqrt((y+8-6)^2+(y+2)^2) )^2+(3sqrt(2))^2}}}

{{{(y+5)^2+(y-1)^2=(y+2)^2+(y+2)^2+18}}}........solve for {{{y}}}

{{{y^2+10y+25 +y^2-2y+1= y^2+4y+4+y^2+4y+4+18}}}

{{{8y+26= 8y+26}}}

{{{y=0}}}

and, then

{{{ x=0+8}}}=>{{{x=8}}}

so {{{Q}}}({{{ 8}}},{{{0}}})


the length of the tangent {{{PQ}}} is: distance between
{{{P}}}({{{6}}};{{{-2}}}) 
{{{Q}}}({{{ 8}}},{{{0}}})

{{{PQ=sqrt((8-6)^2+(0+2) )}}}
{{{PQ=sqrt(2^2+2^2)}}}
{{{PQ=sqrt(2*2^2)}}}
{{{PQ=2sqrt(2)}}}


double check:

{{{(sqrt((8-3)^2+(0-1)^2) )^2=(sqrt((8-6)^2+(0+2)^2) )^2+(3sqrt(2))^2}}}
{{{(5)^2+(-1)^2=(2)^2+(2)^2+18}}}
{{{25+1=4+4+18}}}
{{{26=26}}}


tangent is a line passing through points P({{{6}}};{{{-2}}})  and {{{Q}}}({{{ 8}}},{{{0}}}), so find equation

first find a slope:

{{{m=(0-(-2))/(8-6)}}}
{{{m=2/2}}}
{{{m=1}}}

and use slope point formula

{{{y-y[1]=m(x-x[1])}}}

{{{y-0=1(x-8)}}}

{{{y=x-8}}}-> tangent line

line that passes through {{{P }}}and{{{ M}}} is perpendicular to tangent and it is:


{{{m=(1-(-2))/(3-6)}}}
{{{m=3/-3}}}
{{{m=-1}}}

and use slope point formula

{{{y-y[1]=m(x-x[1])}}}

{{{y-1=-1(x-3)}}}

{{{y-1=-x +3}}}

{{{y=-x +3+1}}}

{{{y=-x +4}}}


{{{drawing( 600, 600, -10, 10, -10, 10,
circle(3,1,.12),locate(3,1,M(3,1)),
circle(6,-2,.12),locate(6,-2,P(6,-2)),
circle(8,0,.12),locate(8,0.5,Q(8,0)),circle(3,1,4.15),


 graph( 600, 600, -10, 10, -10, 10, x-8,-x+4)) }}}