Question 141508
circle with equation: {{{X^2 + Y^2 - 6X - 18Y + 45=0}}}
Tangent touches the circle at point T equation of the tangent is {{{2Y = X}}}
Show algebraically that T has coordinates (6,3).
<pre><font size = 4><b>
Just solve the system

{{{X^2 + Y^2 - 6X - 18Y + 45 =0}}}
{{{2Y = X}}}

by substitution

Substitute 2Y for X in the first:


{{{X^2 + Y^2 - 6X - 18Y + 45 =0}}}

{{{(2Y)^2 + Y^2 - 6(2Y) - 18Y + 45 =0}}}

{{{4Y^2+Y^2-12Y-18Y+45=0}}}

{{{ 5Y^2-30Y+45=0 }}}

Divide every term by 5

{{{Y^2-6Y+9=0}}}

Factor:

(Y-3)(Y-3)=0

{{{Y-3=0}}} gives {{{Y=3}}}
{{{Y-3=0}}} gives {{{Y=3}}}

[The fact that we get only one solution tells us
that the line touches the circle in only one
point. So this proves that it is indeed tangent]

Now to find {{{X}}}

Substitute {{{3}}} for {{{Y}}} in

{{{2Y = X}}}

{{{2(3)=X}}}
6=X

So the point of tangency is (6,3)

Edwin</pre>