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

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

• Please use MathJax – Rishi Sep 19 '19 at 6:03

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 $$r=|x_c|$$.

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.

• yes, from your pic, we can just calculate, BQ = 2AQ right? – Lifeforbetter Sep 23 '19 at 14:42
• @Lifeforbetter: $BQ=\sqrt{2} AQ$ – Jack D'Aurizio Sep 23 '19 at 14:44
• oh, i meant i project the center of circle 1 as A', so BQ = 2A'Q. can i? – Lifeforbetter Sep 23 '19 at 14:48
• can i take point 1,3 to be parallel to B as A', so, A'Q = $(1-0)^2 + (3-b)^2 = distance^2$ BQ = \$(2-0)^2 + (4-y)^2 = D^2 – Lifeforbetter Sep 23 '19 at 14:54
• what is the educated guess you talked about? – Lifeforbetter Sep 23 '19 at 14:55

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: $$(x-1)^2+(y-4)^2=1$$ $$(x-5)^2+y^2=25$$

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.