A line through the point P(8, -7) is a tangent to the circle C at the point T. Find T 
Circle C equation $(x+5)^2+(y-9)^2=25$
  A line through the point P(8, -7) is a tangent to the circle C at the point T.
  Find T.

I tried simultaneous equations:
1. $(x+5)^2+(y-9)^2=25$
2. $y = m(x-8)-7$
3. differentiate circle: $\dfrac {dy} {dx} = \dfrac{-(x+5)} {y-9}$
where x, y  represent co-ordinates of T  
But it does not seem to be getting anywhere.
By using first 3 equations I get horrible mix of $x, x^2, x^3, y^2, y, xy$ 
Many thanks in advance
 A: plug in $ m= \dfrac {dy} {dx} = \dfrac{-(x+5)} {y-9}$  you get
$$y = \dfrac{-(x+5)(x-8)} {y-9}-7 \Rightarrow (y+7)(y-9) = -(x+5)(x-8) 
  $$
so your answer will be the solution of the following system ..
$\begin{array}{r c l} (y+7)(y-9) + (x+5)(x-8) &=&0 \\ (x+5)^2+(y-9)^2 &=&25 \end{array}$
you just have to expand both equations and subtract them to cancel all the quadratic terms and you will be left with the equation of a line.
The solutions to your problem will be the points of intersection between that line and the circle 
$(x+5)^2+(y-9)^2 =25 $
A: Draw the circle whose diameter is the line segment between $(-5\mid 9)$ and $(8\mid -7)$. Its two intersections with the circle you supplied are the points you seek. (What do we know about an angle inscribed in a semicircle whose legs pass through both ends of the diameter?)   
$(-8\mid 5)$ and $\left(-\frac8{17}\mid\frac{189}{17}\right)$.
A: Let the co-ordinates of the point $T\equiv(a, b)$ then lines PT & CT are normal to each other then $$\left(\frac{b-(-7)}{a-8}\right)\times \left(\frac{b-9}{a+5}\right)=-1$$ $$a^2+b^2-3a-2b-103=0   \tag 1$$
now, satisfy the equation of circle by point $T\equiv(a, b)$, we get
$$(a+5)^2+(b-9)^2=25$$ $$a^2+b^2+10a-18b+81=0   \tag 2$$ subtracting eq(1) from (2), we get $$13a-16b+184=0  $$$$=> b=\frac{13a+184}{16}  \tag3$$ 
Now, in right $\Delta PTC$, we have $$(PT)^2=(CP)^2-(CT)^2$$ $$ (a-8)^2+(b+7)^2=(8+5)^2+(-7-9)^2-(5)^2$$ Substituting the value of $b$ from eq (3) in above expression, we get a quadratic equation in terms of $a$ as follows
$$(a-8)^2+\left(\frac{13a+184}{16}+7 \right)^2=400$$ $$17a^2+144a+64=0$$ By solving above quadratic equation, we get $a=-8 => b=5 $ & $a=\frac{-8}{17} => b=\frac{189}{17}$ thus we get two points of tangency on the circle as follows $$T\equiv\left(-8, 5\right)$$ & $$T\equiv\left(\frac{-8}{17}, \frac{189}{17}\right)$$
A: The equation of tangent to a circle $ x^2+y^2+2gx+2fy+c=0 $   at a point $(x_1,y_1)$ is $ xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0$. 
using this the equation of required line will be $xx_1+yy_1+5(x+x_1)-9(y+y_1)+81=0$  and this line passes through $(8,-7)$, which gives you a equation in $(x_1,y_1)$.
you can solve this directly with the equation of circle to get the point $(x_1,y_1)$
or, since this line is a tangent the distance of this line from the center $(-5,9)$ will be equal to radius ie $(x_1+5)^2 + (y_1-9)^2=25$. and hence you can get the coordinates of $(x_1,y_1)$.
