Determine the equations of the circles

Question

EXERCISE: Determine the equations of the circles that pass through the points $(2, 3)$ and $(3, 6)$, and are tangent to the line $2x + y − 2 = 0.$

My idea was, first, to notice that the points $(2,3)$ and $(3,6)$ belong to the circle, then substitute in the general equation the equation of a circle.
Thus I arrived at the equation of the possible centers of the circumference.
Then, it remained to look for the circles with a center satisfying the equation that I found, and that were tangent to the line $2x+y-2=0,$ so I calculated the distance between a possible center and the tangent and said that this was equal to the radius.
But then I didn't know what to do. I hope someone can help me to solve it

Answer 1

The radius will be $$r=d(l, C) =\frac{|2c_1+c_2-2|}{\sqrt{5}}, $$ where $C=(c_1, c_2)$ is the center. Since the center verifices that $2c_1+6c_2-32=0$, it implies that $c_1=16-3c_2$, so $$r=\frac{|32-6c_2+c_2-2|}{\sqrt{5}}=\frac{|30-5c_2|}{\sqrt{5}}=\sqrt{5}|6-c_2|.$$ Can you conclude from here?

Answer 2

I don't undertand how the accepted answer arrives at the solution (well at least one of them), so I will explain how I did it. We are given two points (2,3) and (3,6), we also know a point of the line $\large y = -2x + 2$ is part of the circle, let's call it (x,y) and let's call the center of the circle (h,k), using the equation of a circle we get the following system of equations: $$\large \begin{cases}(x-h)^2+(y-k)^2=r^2\\(2-h)^2+(3-k)^2=r^2\\(3-h)^2+(6-k)^2=r^2\end{cases}$$ Now we need to turn this system of equations into one in 3 variables, to do this we'll express x and y in terms of h and k using relationships between these variables.
We know that (x,y) lie on a line which is perpendicular to $\large y=-2x+2$ and passes through (h,k), so: $$\large y= \frac{1}{2}(x-h)+k$$ Using the equation of the tangent line we express y in terms of x and substitute into the perpendicular line equation: $$\large -2x+2=\frac{1}{2}(x-h)+k$$ And now we solve for x: $$\large x=\frac{h-2k+4}{5}$$ So, in resume, we can express y in terms of x, and x in terms of h and k, and the system of equations becomes: $$\large \begin{cases} (\frac{h-2k+4}{5}-h)^2+(-2\frac{h-2k+4}{5}+2-k)^2=r^2\\ (2-h)^2+(3-k)^2=r^2\\ (3-h)^2+(6-k)^2=r^2 \end{cases}$$ Plugging this system into an equation solver (because you don't want to lose your sanity) gives: $$\large(h,k,r)=(1,5,\sqrt{5})$$ $$\large(h,k,r)=(13,1,5\sqrt{5})$$ So the circles' equations are: $$\large (x-1)^2+(y-5)^2=5$$ $$\large (x-13)^2+(y-1)^2=125$$ diagram by courtesy of PM 2Ring

Answer 3

Label the given points $A(2,3), B(3,6)$ and the line $p: 2x+y-2=0.$
The center of every convenient circle lies on the bisector $b$ of the segment $AB,$ its equation is $$b: x+3y-16=0$$ and parametric equations \begin{equation} {b:\bigg\{\begin{matrix}x = & 3t -16 \\ y = & t \end{matrix}} \qquad t\in\mathbb{R}. \end{equation} We are searching for all $\;t\in\mathbb{R}\;$ such that the point $S_t$ is equidistant to $A,B$ and $p.$ It is clear that $|AS_t|=|BS_t|.$ After plugging into $$dist(S_t,p)=|BS_t|$$ and some simple algebra, we obtain $$\sqrt5 \cdot|6-t|=\sqrt{(13-3t)^2+(t-6)^2}$$ Squaring and simplifying leads to a quadratic equation $$t^2-6t+5=0.$$ The rest is straightforward.