Orthogonal circles

Question

What is the equation of the circle that is orthogonal to the circles $x^2 + y^2 - 8x +5 =0$ and $x^2 + y^2 +6x +5 = 0$ and passes through the point $(3,4)$? I've spent hours trying to figure this out - help please

Answer 1

Just have a look at the following Wikipedia page. Two circles with equations: $$ x^2+y^2+2g x+2f y + c=0 $$ $$ x^2+y^2+2g' x+2f' y + c'=0 $$ are orthogonal if: $$(\clubsuit)\quad 2 gg'+2ff' = c+c', $$ hence by assuming that the equation of our circle is $x^2+y^2+2ax+2by+c=0$ we have: $$ 6a = c+5 = -8a, $$ so $a=0$, $c=-5$ and we just need to impose that our circle goes through $(3,4)$ to find $b$, too.

It is not difficult to prove that $(\clubsuit)$ is equivalent to: the center of the third circle lies on the radical axis of the first two circles.

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As an alternative, you may just find the circular inverses of $(3,4)$ with respect to the two initial circles. The circle through the last two points and $(3,4)$ is the solution:

Answer 2

you have $$C_1:(x-4)^2+y^2=11,\quad C_2:(x+3)^2+y^2= 4$$ we will look for a point $(a,0)$ on the $x$-axis so that the tangents to the circle have the same length.

we have $$(4-a)^2 -11=(a+3)^2-4 \to-8a+5=6a+5\to a = 0.$$

we will use the fact that if a circle is orthogonal to both $C_1, C_2,$ then the $x$-cdt of the center is $0.$

let the the center of $C$ that is orthogonal to both $C_1, C_2$ and goes through $(3,4)$ be $(0,b)$.

$$x^2 + (y-b)^2 = b^2 + 9-4 \to x^2+y^2 - 2by=5$$ making this go through $(3,4)$ requires $$3^2 + 4^2-8b=5\to b= \frac 52 $$ andn the circle is $$x^2 + y^2-5y=5. $$

Answer 3

Given two circles, $(c_a,r_a)$ and $(c_b,r_b)$, the locus of points, $x$, that have equal tangents to both circles can be computed as follows.

As can be seen above, $$ |x-c_a|^2-r_a^2=|x-c_b|^2-r_b^2\tag{1} $$ which is equivalent to $$ \left(x-\frac{c_b+c_a}2\right)\cdot(c_b-c_a)=-\frac12(r_b^2-r_a^2)\tag{2} $$ The $x$ that satisfy $(2)$ form a line perpendicular to $c_b-c_a$ that passes through the point $$ \begin{align} &\frac{c_b+c_a}2-\frac12\frac{r_b^2-r_a^2}{|c_b-c_a|^2}(c_b-c_a)\\[6pt] &=\frac{c_a\left(|c_b-c_a|^2+r_b^2-r_a^2\right) +c_b\left(|c_b-c_a|^2+r_a^2-r_b^2\right)}{2|c_b-c_a|^2}\tag{3} \end{align} $$ As can also be seen in the diagram, each point on this line is also the center of a circle that is orthogonal to both $(c_a,r_a)$ and $(c_b,r_b)$.

Furthermore, passing through a point is the same as being orthogonal to a circle of radius $0$ centered at that point.

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The circles in the problem are $$ \left\{\left((4,0),\sqrt{11}\right),\left((-3,0),2\right),\left((3,4),0\right)\right\}\tag{4} $$ Using $(3)$, the locus of points for the first two circles is $$ (0,0)+(0,1)t_1\tag{5} $$ and the locus of points for the first and third circles is $$ \left(\tfrac{54}{17},\tfrac{56}{17}\right)+(4,1)t_2\tag{6} $$ The intersection of $(5)$ and $(6)$ is where $$ \left(\left(\tfrac{54}{17},\tfrac{56}{17}\right)+(4,1)t_2\right)-\left((0,0)+(0,1)t_1\right)=(0,0)\tag{7} $$ Take the dot product of $(7)$ with $(-1,4)$ to get $t_1=\tfrac52$. Therefore the intersection is $\left(0,\tfrac52\right)$.

We can check that for all three circles, $$ \left|\,c-\left(0,\tfrac52\right)\,\right|^2-r^2=\tfrac{45}4\tag{8} $$ and get that the circle orthogonal to all three circles is $$ \left(\left(0,\tfrac52\right),\tfrac{3\sqrt5}2\right)\tag{9} $$

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The equation of the circle in $(9)$ is $$ \bbox[5px,border:2px solid #C0A000]{x^2+\left(y-\tfrac52\right)^2=\tfrac{45}4}\tag{10} $$ which can also be written as $$ \bbox[5px,border:2px solid #C0A000]{x^2+y^2-5y=5}\tag{11} $$