Given that equation of a circle $C$ is $x^2 +y^2 -10x -8y +18=0$
Find the equations of the tangents to the circle $C$ which are parallel to line $y-x+5=0$
(Originally this was not a differentiation question,but I found out that if I restricted myself to pre-calculus,I would have to use discriminant and then the quadratic formula.)
$x^2 +y^2 -10x -8y +18=0$
By differentiating with respect to $x$,
$2x +2y y' -10 -8y'=0$
Hence,
$y'=\frac{-2x+10}{2y-8}$
As the gradient is 1,
$2y-8=-2x+10$
$y=-x+9$
That's where I got stuck.As the gradient is $1$,why does last equation has a gradient of $-1$?Where did I go wrong?Lastly,is there any other easier way ?
Edit:
Subst. $y=-x+9$ into $C$
$x^2 + (9-x)^2 -10x-8(9-x)+18=0$
$2x^2 +81-18x -10x -72+8x+18=0$
$2x^2 -20x -27=0$
This would give me
$x=\frac {10+\sqrt(154)}{2} $or $x=\frac {10-\sqrt(154)}{2}$
Hence,$y=\frac {8+-\sqrt(154)}{2}$
So,the equation is $y=x-1+-\sqrt(154)$
However when I did it the pre-calculus way,the answer is $ y=x-1+- \sqrt(46) $
Best Answer
You have to complete your work.
Take $y=-x+9$ in the equation of the circle that becomes: $$ 2x^2-20x+27=0 $$ solve for $x$ and find the two roots $x_1$ and $x_2$. Find $y_1=-x_1+9$ and $y_2=-x_2+9$ . The points $(x_1,y_1)$ and $8x_2,y_2)$ are the points of tangency. Now write the equations of the two lines of slope $m=1$ through these two points.
Note that your solution is wrong because you have a $-27$ in the equation. The correct solution for $x$ is: $$ x=\frac{10\pm \sqrt{46}}{2} $$
The other method, without derivative, is to search a line of equation $y=x+q$ that is tangent to the circle. This means that the system $$ \begin{cases} y=x+q\\ x^2+y^2-10x-8y+18=0 \end{cases} $$ has only one solution. So, substitute $y$ in the second equation so that you have a second degree equation in $x$ with a parameter $q$ and find the value of the parameter $q$ such that the discriminat of this equation is null $\Delta(q)=0$. Since $\Delta(q)$ is a second degree polynomial in $q$ you find two values for $q$ that are the intercepts of the the two tangent lines.
Another method is to intersect the line orthogonal to the $y-x+5=0$ and passing through the center of the circle with the circle. The common points are the points of tangency.