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.
Best Answer
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.