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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.

Aditya
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2 Answers2

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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.

Menezio
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  • So B and C lie on the circle. But I still have to find the coordinates, which would, require me to solve the two equations – Aditya Mar 30 '20 at 09:16
  • Well the computation is easier beacuse in the equation of degree $2$ you know a solution that is the point $A$. Anyway there are an other way: Take the rotation by $\pm 90^\circ$ with center the center of your circle and apply it to the point $A$. I think that are the shortest ways; they are not so much tedius – Menezio Mar 30 '20 at 09:29
  • @Aditya I included more details – Menezio Mar 30 '20 at 09:59
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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.

amd
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