There are two straight lines passing through the point A(2,0) which make an angle of 45 deg with the tangent at $A$... 
There are two straight lines passing through the point $A(2,0)$ intersecting the tangent from $A$ of the circle $x^2+y^2+4x-6y-12=0$ under $45^{\circ}$. Find the equation of the circles with radius $3$ units each, centred on these straight lines at a distance of $5\sqrt 2$ from $A$

The equation of the tangent at $A$ is $4x-3y-8=0$
Doing a fair bit of calculation, the equations of the two straight lines are 
$$x-7y-2=0$$
$$7x+y-14=0$$
How do I find the centre of the circles? 
Two ways came to my find 
The Symmetric form
The formula $$\frac{x-x_1}{\cos \theta}=\frac{y-y_1}{\sin \theta}=r$$
Solving equations
Let the centre be $(h,k)$
Then 
$$(h-2)^2 + k^2=50$$
And solving this with 
the two obtained lines 
Both methods are horribly tedious to solve. I am sure there is a more efficient way to solve. 
 A: Observe that the two straight lines intersect the circles in two points $B,C$. Now, thanks to ortogonality of straight lines, you have $B\hat A C = 90^\circ$. So the points $A,B,C$ form a part of the square inscribed in the circle: moreover the diagonal of this square is the diameter of the circle that is $10$. Thanks to Pythagorean theorem you have $\overline{AC} = \overline{AB} = 5\sqrt{2}$ so the points $B$ and $C$ are what you required in your question. To find $B$ and $C$ we can rotate the point $A$ by $90^\circ$ around $O=(-2,3)$ center of your circle. The rotation is given by the formula (center is $(a,b)$ and the angle $\alpha$):
\begin{cases} 
x' = (x-a)\cos\alpha - (y-b)\sin\alpha +a\\
y' = (x-a)\sin\alpha + (y-b)\cos\alpha +b\\
\end{cases}
So you have ($\alpha = 90^\circ$, $O$ the center)
\begin{cases} 
x' =  - (y-3) -2\\
y' = (x+2)  +3\\
\end{cases}
and applying it to $A$ you have $B= (1,7)$. Doing the same thing with $\alpha = -90^\circ$ you have $C=(-5,-1)$.
For the other two points we can use also rotation: take $B$ and $C$ and rotate them around $A$ for $180^\circ$. Again you can use the above formula that become ($A$ is the center of rotation and $\alpha = 180^\circ$)
\begin{cases} 
x' = 4-x\\
y' = -y\\
\end{cases}
And you finally obtain $B'=(3,-7)$ and $C'=(9,1)$.
Summarizing the four points are $B=(1,7)$, $B'=(3,-7)$, $C=(-5,-1)$, $C'=(9,1)$ and we did not resolve any equation.
A: The two 45-degree lines are the angle bisectors of the tangent and normal through $A$. You’ve already worked out an equation $4x-3y-8=0$ of the tangent, and a small amount of additional work yields $3x+4y-6=0$ for the normal. The angle bisectors of these two lines are $(4x-3y-8)\pm(3x+4y-6)=0$. The circle centers are the intersections of these two lines with the circle of radius $5\sqrt2$ centered at $A$.  
Alternatively, observe that the 45-degree lines to the tangent form an inscribed angle of 90 degrees in the given circle at $A$, so their other intersections with the circle are endpoints of a diameter. The radius of this circle happens to be equal to $5$, so two of the points that you seek are in fact these diameter endpoints. This diameter is parallel to the tangent at $A$, so it shouldn’t be very hard to find these points knowing the equation of this tangent and the center of the circle. The other two are then obtained by reflection in the tangent at $A$. This is essentially Menezio’s solution.
