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
Best Answer
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.