Finding the radius of the circles given rules

Question

There are two circles that go through the coordinates (1,3) and (2,4) and are tangent to the y-axis. Let $a$ and $b$ be the radius of those circles, what is the value of $a\cdot b$?

My approach:

The question already tells that there are only two circles that attend to these demands. I made a system of 2 equations using the circle equations:

$(x-x_0)^2+(y-y_0)^2=r^2$

Putting those coordinates from the question, it becomes a system:

$(1-x_0)^2+(3-y_0)^2=r^2$

$(2-x_0)^2+(4-y_0)^2=r^2$

Resolving this system by computating the squares and subtracting one from another, I get:

$x_0+y_0=5$

But this gives all the origins that go through those coordinates including the ones that pass through the y-axis. I would like to know if the approach is correct, or else what am I missing.

By trial and error, I could find that those 2 circles are centered at $(1,4)$ and $(5,0)$, which give the respective radius of $1$ and $5$, thus the answer is $a\cdot b = 5$. However, is there another method of solving this?

Answer 1

You've set this up correctly, but you haven't used the fact that the circle is tangent to the $y$-axis. When the circle is tangent to the $y$-axis, the left-most or right-most point of the circle will be on the $y$-axis. In other words, $(0,y_0)$ must be a point on the circle. Putting this together with what you already have, you get:

\begin{align} (1-x_0)^2+(3-y_0)^2&=r^2\\ (2-x_0)^2+(4-y_0)^2&=r^2\\ (0-x_0)^2+(y_0-y_0)^2&=r^2. \end{align}

Expanding everything, we get \begin{align} 1-2x_0+x_0^2+9-6y_0+y_0^2&=r^2\\ 4-4x_0+x_0^2+16-8y_0+y_0^2&=r^2\\ x_0^2&=r^2 \end{align}

Substituting the third equation into the first two gives \begin{align} 1-2x_0+x_0^2+9-6y_0+y_0^2&=x_0^2\\ 4-4x_0+x_0^2+16-8y_0+y_0^2&=x_0^2 \end{align}

Simplifying, we get \begin{align} 1-2x_0+9-6y_0+y_0^2&=0\\ 4-4x_0+16-8y_0+y_0^2&=0 \end{align}

Solving each equation for $x_0$ we get, \begin{align} \frac{1}{2}(1+9-6y_0+y_0^2)&=x_0\\ \frac{1}{4}(4+16-8y_0+y_0^2)&=x_0 \end{align}

Since both equations equal each $x_0$, the LHSs are equal to each other and we get $$ \frac{1}{2}(1+9-6y_0+y_0^2)=\frac{1}{4}(4+16-8y_0+y_0^2). $$ Can you take it from here?

Answer 2

Both given points have positive abscissa, so the circles must be tangent to the Y axis from the positive side. Then the radius of a circle equals the $x$ coordinate of its center and the equation is $$(x-r)^2+(y-y_0)^2=r^2.$$ When we substitute the given coordinates we get $$\begin{cases} (1-r)^2+(3-y_0)^2=r^2 \\ (2-r)^2+(4-y_0)^2=r^2 \end{cases}$$ Expanding a bit: $$\begin{cases} 1-2r +r^2+(3-y_0)^2=r^2 \\ 4-4r +r^2+(4-y_0)^2=r^2 \end{cases}$$ After cancelling $r^2$: $$\begin{cases} (3-y_0)^2 = 2r-1 \\ (4-y_0)^2 = 4r-4 \end{cases}\tag{*}$$ so $$(4-y_0)^2 - 2(3-y_0)^2 = -2$$ Solve for $y_0$, plug the values to $(*)$ and you'll get possible values for $r$.

Answer 3

After giving the answer I got another idea: when the problem involves quadratic equations and the task is to find a product of the two roots, is seems to be asking for Vieta's formulas. How could we make use of it?

First we need an equation with $r$ as unknown variable.
Let's see… Can we define the circle with $r$?

The circle passes through two given points $A(1,3)$ and $B(2,4)$. That implies the center lies on the bisector of the line segment $AB$.
The bisector passes the line segment's midpoint $M(1.5,3.5)$ and has a slope $-1$, because that's the number opposite to a reciprocal of the line $AB$ slope. Hence the equation of the bisector is $$y = 5-x$$ We need a circle tangent to the Y axis, so its center $S$ must be at the distance $r$ from the axis. Then it has coordinates $(r,5-r)$ and its distance from $A$ is $$SA = \sqrt{(r-1)^2+((5-r)-3)^2}$$ We want the circle centered at $S$ to pass through $A$, so we require $$SA=r$$ That is $$\sqrt{(r-1)^2+(2-r)^2} = r$$ $$(r-1)^2+(2-r)^2 = r^2$$ $$2r^2 -6r + 5 = r^2$$ $$\color{red}1r^2 -6r + \color{red}5 = 0$$ And, from the Vieta's formula, the product of roots $$\boxed{ r_1\cdot r_2=\frac{\color{red}5}{\color{red}1} = 5}$$