2 circles go through points $(1,3)$ and $(2,4)$ tangent to $y$ axis

Question

Find radii of both circles.

Center for circle 1 is $(a_1,b_1)$. Tangent at $y$ axis at $(0,k)$. Radius of circle 1 is $r_1^2 = a_1^2 + (b_1-k)^2$.

Center for circle 2 is $(a_2,b_2)$. Tangent at $y$ axis at $(0,h)$. Radius of circle 2 is $r_2^2 = a_2^2 + (b_2-h)^2$

Mid point of $(1,3)$ and $(2,4)$ is $(1.5,3.5)$ Line goes through mid point $y_1=x+2$ Line perpendicular through y1 and goes through both centers is $ y_2=-x+5$

From substitute $(1,3)$ and $(2,4)$ to equation of circle 1 I get $a_1+b_1 =2$. And from $ y_2$, i get $a_1+b_1 = 5$

Despite all information i can find, i still get stuck to find the radii. and find the $a_1,b_1$ or $a_2,b_2$ to at least get radii.

How to get radii using the line $y_2$, or circle equation 1 or other way? What am i missing? Please your clarification

Answer 1

Let's start with the equation of the circle: $$(x-x_c)^2+(y-y_c)^2=r^2$$ Then we can plug in the two points on the circle: $$(1-x_c)^2+(3-y_c)^2=r^2\\(2-x_c)^2+(4-y_c)^2=r^2$$ We also know that the circle is tangent to $y$ axis. That means that the intersection of the circle with the $y$ axis is at the same $y$ as $y_c$ (the radius is horizontal): $$(0-x_c)^2+(y_c-y_c)^2=r^2$$ or $$x_c^2=r^2$$ Now plug this into the first two and you get $$1-2x_c+x_c^2+9-6y_c+y_c^2=x_c^2\\4-4x_c+x_c^2+16-8y_c+y_c^2=x_c^2$$ We rewrite the equations as $$2x_c=10-6y_c+y_c^2\\4x_c=20-8y_c+y_c^2$$ Substituting the first equation into the second we get: $$20-12y_c+2y_c^2=20-8y_c+y_c^2$$ or $$y_c^2-4y_c=0$$ The solutions are $y_c=0$ and $y_c=4$. You can now get the corresponding $x_c$ and $r=|x_c|$.

Answer 2

The radius of a tangent circle to $y$ axis by center of $(x_0,y_0)$ is $|x_0|$. So if you continue you must choos a point $(x_0,y_0)$ such that: $$(x_0-1)^2+(y_0-3)^2=x_0^2=(x_0-2)^2+(y_0-4)^2$$ That is gives you two answer, $(1,4) \& (5,0)$, so the circles equations are as follow: $$(x-1)^2+(y-4)^2=1$$ $$(x-5)^2+y^2=25$$

Answer 3

Three points $(1,3), (0,k+\epsilon), (0,k-\epsilon)$, determine a circle. Letting $\epsilon\to 0$ makes this tangent to the $y$-axis at $(0,k)$. Substituting $(2,4)$ gives the possible values for $k$.

In M2

R=QQ[k,e]
S=R[x,y]
factor determinant matrix {{x^2+y^2,x,y,1},{10,1,3,1},{(k+e)^2,0,k+e,1},{(k-e)^2,0,k-e,1}}
--(e)*(x^2+y^2+(e^2-k^2+6*k-10)*x-2*k*y-e^2+k^2)*(2)

substitute $e=0, (x,y)=(2,4)$ into the second factor to get $$-k(k - 4)$$ So $k=0\vee k=4$, or $$x^2+y^2-10x$$ and $$x^2+y^2-2x-8y+16.$$

The radii are 1 and 5.

Answer 4

Let us consider the centers $C$ and $D$ of these circles. Both circles have to go through $A,B$, so both $C$ and $D$ lie on the perpendicular bisector of $AB$. The distance of $C$ from $A$ equals the distance of $C$ from the $y$-axis. The distance of $D$ from $A$ equals the distance of $D$ from the $y$-axis, so both $C$ and $D$ lie on the parabola having the $y$-axis as directrix and the point $A$ as focus. It follows that the problem is solved by intersecting a line ($y=5-x$) and a parabola ($x=\frac{1+y^2}{2}$).

From $C(1,4)$ and $D(5,0)$ it follows that $r_C=1$ and $r_D=5$. These solutions can be found by educated guess, too: $C$ is trivially the center of a circle through $A$ and $B$ which is tangent to the $y$-axis. $D$ clearly works as a center since $3,4,5$ is a Pythagorean triple.

Alternative approach: let $P$ be the intersection of the $AB$-line with the $y$-axis. Let $Q,R$ be the tangency points of our circles with the $y$-axis. By considering the powers of $P$ with respect to our circles we have $$ PQ^2 = PR^2 = PA\cdot PB = \sqrt{2}\cdot 2\sqrt{2} = 4 $$ so $PQ=PR=2$, the locations of $P,Q$ are straightforward to be found and the locations of $C,D$ too.