A line through the point P(8, -7) is a tangent to the circle C at the point T. Find the length of PT. 
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 the length of PT.

The question itself is easy when using pythagoras,
But I cannot understand some of the alternative methods written in the solution paper:
"Substitutes (8, -7) into circle equation so $PT^2 = 8^2 + (−7)^2 +10×8−18×(−7) + 81$"
what is this method?
And further:
"A significant number “found” the coordinates of one point of contact T, (–8, 5), often just stating it. A number then tried to explain the solution with 3 4 5 triangles, to which the points concerned lent themselves. It was noticeable that the coordinates of the other point of contact were never found in this way."
How do I find the co-ordinate of T is (-8,5)? given that some people just stated it there must be a quick way.
Only method I can think of is horrible simultaneous equations:
differentiate the circle equation and equate that with line equation $y + 7 = \dfrac {9-T_x} {-5-T_y}(x-8) $
But I have problem finding gradient alone in this case because there are too many unknowns.
Many thanks in advance. 
 A: Calculate Power of the circle by shifting P to origin and the circle also accordingly for same relative position.
$$ ( x + 5 + 8)^2 + ( y -9 -7)^2 = 25 $$
Tangent length is square root of Power = $ \sqrt{13^2 + 16^2 -25} =20. $ by the Pole/Polar method.
A: HINT: make the ansatz $$y=mx+n$$ and for $x=8,y=-7$ we get
$$y=m(x-8)-7$$ and now plug this equation in 
$$(x+5)^2+(y-9)^2=25$$
then you must calculate $$m$$ since the line must be a tangent line 
A: If the center of the circle is point C then the triangle $\triangle CTP$ is a right triangle. So if the radius of the circle is $r={CT}$ and the distance from the center to the point is $d=CP$ then the required distance is $$t=\sqrt{d^2-r^2}$$
A: Circle having equation: $(x+5)^2+(y-9)^2=25$ has center $C\equiv(-5, 9)$ & a radius $5$ units. Then the distance of the external point $P\equiv(8, -7)$ from the center $C\equiv(-5, 9)$ is calculated as $$PC=\sqrt{(8-(-5))^2+(-7-9)^2}=\sqrt{425}$$ Hence, in right $\Delta PTC$, we have $$PT=\sqrt{(PC)^2-(CT)^2}=\sqrt{(\sqrt{425})^2-(5)^2}=\sqrt{425-25}=20 \space units$$
