Question 1129332
a. Show that the point Q({{{6}}},{{{2}}}) lies on the circle 

{{{x^2+y^2-4x+2y-20}}}
{{{6^2+2^2-4*6+2*2-20}}}
{{{36+4-24+4-20}}}
{{{44-44}}}
{{{0}}}=>the point Q({{{6}}},{{{2}}}) lies on the circle


b. Find the equation of tangent to the circle at point Q 


As a tangent is a straight line it is described by an equation in the form 
{{{ y - y[1] = m(x -x[1])}}}. 

You need both a {{{point}}} and the {{{slope}}} to find its equation.

You are usually given the point - it's where the tangent meets the circle.

To find the {{{slope}}} use the fact that the tangent is {{{perpendicular}}} to the {{{radius}}} from the point it meets the circle.

Work out the slope of the {{{radius}}} (r) at the point the tangent meets the circle. Then use the equation  {{{ m_tg = - 1/m_r}}} to find the gradient of the tangent.

{{{(x^2-4x)+(y^2+2y)=20}}}...if you complete squares, you will see that the center of the circle is  ({{{2}}},{{{-1}}})

{{{(x^2-4x+b^2)-b^2+(y^2+2y+b^2)-b^2=20}}}
{{{(x^2-4x+2^2)-2^2+(y^2+2y+1^2)-b^2=20}}}
{{{(x-2)^2)-4+(y+1)^2)-1=20}}}
{{{(x-2)^2)+(y+1)^2)=25}}}

Q({{{6}}},{{{2}}}) and  ({{{2}}},{{{-1}}})=>

 the slope of the radius 

{{{m_r=(-1-2)/(2-6)=-3/-4=3/4}}}

the slope of the tangent is {{{m_tg=-1/(3/4)=-4/3}}}

and equation af the tangent is:


Q({{{6}}},{{{2}}}) ,{{{m_tg=-4/3}}}

 {{{y - 2 = -(4/3)(x -6)}}}
{{{y  = -(4/3)x -6(-4/3)+2}}}
{{{y  = -(4/3)x +8+2}}}
{{{y  = -(4/3)x +10}}}


{{{ drawing( 600, 600, -15, 15, -15, 15,
circle(6,2,.12),locate(6,2,Q(6,2)),circle(2,-1,.12),locate(2,-1,C(2,-1)),
 graph( 600, 600, -15, 15, -15, 15,-(4/3)x +10,sqrt(25-(x-2)^2)-1,  -sqrt(25-(x-2)^2)-1)) }}}