Now, if I suppose that the equation is looks like $$(x-p)^2 + (y-q)^2 = r^2$$
Since the circle touches another circle whose radius is $1$, we can write that $q=1+r$. However, right here is where I'm stuck. I don't know how to express the other terms, and I'm not sure what I need the point $M(4,3)$ for. Could anyone help?
Since the circle touches the $y$-axis we have $p=r>0$. Since the circle passes through $(4,3)$ we have
$$(4-r)^2+(3-q)^2=r^2.$$
Suppose it touches the other circle at point $(a,b),$ which must be a point on both circles:
$$(a-2)^2+b^2=1.$$
$$(a-r)^2+(b-q)^2=r^2.$$
Finally, tangency between two circles at the point $(a,b)$ gives an equal slopes condition:
$$ {a-r \over b-q}={a-2\over b}.$$
Four equations. Four unknowns. Solving gives four solutions:
$a=2,b=1,r=2,q=3.$
$a=74/25,b=-7/25,r=74,q=-21.$
$a={2\over 73}(53 \pm 6\sqrt{3}),b=-{1 \over 73}(15 \pm 32\sqrt{3}),r={2 \over 3}(7 \mp 2\sqrt{3}),q=-{1 \over 3}(3 \mp 4\sqrt{3}).$
Let $(p,q)$ be the centre of the required circle, and $r$ be its radius.
As the required circle touches the $y$-axis, $p=r$. (Rejecting the other case when $p = -r$)
As the required circle passes through $M(4,3)$,
$$\begin{align*} (4-p)^2 + (3-q)^2 &= r^2\\ 16 - 8r + r^2 + 9 - 6q + q^2 &= r^2\\ q^2 - 6q + 25 &= 8r \tag1 \end{align*}$$
Let $C$ be the given circle $(x-2)^2 + y^2 = 1$. If the required circle touches $C$ externally, then the distance between their centres is $r+1$:
$$\begin{align*} (p-2)^2 + (q-0)^2 &= (r+1)^2\\ r^2 - 4r + 4 + q^2 &= r^2 + 2r + 1\\ q^2 + 3 &= 6r \tag 2 \end{align*}$$
Eliminating the $r$ in $(1)$,
$$\begin{align*} q^2 - 6q + 25 &= \frac 43(q^2 + 3) \\ 3q^2 - 18q + 75 &= 4q^2 + 12 \\ 0 &= q^2 + 18q - 63\\ 0 &= (q+21)(q-3)\\ \end{align*}$$
The two circles in this case are:
$q=3$, $r =\frac{3^2+3}6 = 2$, so the circle is $$C_1: (x-2)^2+(y-3)^2 = 2^2;$$
or $q=-21$, $r =\frac{(-21)^2+3}6 = 74$, so the circle is
$$C_2: (x-74)^2+(y+21)^2 = 74^2.$$
In the other case, when the required circle touches $C$ internally, then the distance between their centres is $r-1$:
$$\begin{align*} (p-2)^2 + (q-0)^2 &= (r-1)^2\\ r^2 - 4r + 4 + q^2 &= r^2 - 2r + 1\\ q^2 + 3 &= 2r \tag 3 \end{align*}$$
Eliminating the $r$ in $(1)$,
$$\begin{align*} q^2 - 6q + 25 &= 4(q^2 + 3)\\ 0 &= 3q^2 + 6q - 13\\ q &= \frac{-6 \pm \sqrt{6^2 - 4\cdot 3(-13)}}{2\cdot3}\\ &= -1 \pm \frac{4}{\sqrt3}\\ q^2 &= 1 \pm 2(-1)\cdot \frac{4}{\sqrt 3} + \frac{4^2}3\\ &= \frac{19}{3} \mp 2\cdot \frac{4}{\sqrt 3}\\ r &= \frac{q^2+3}{2} = \frac{14}{3} \mp \frac 4{\sqrt 3} \end{align*}$$
The two circles in this case are
Taking $+$ in $q$ and $-$ in $r$, the circle is
$$C_3: \left[x-\left(\frac{14}{3} - \frac 4{\sqrt 3}\right)\right]^2 + \left[y-\left(-1 + \frac{4}{\sqrt3}\right)\right]^2 = \left(\frac{14}{3} - \frac 4{\sqrt 3}\right) ^2;$$
or taking $-$ in $q$ and $+$ in $r$, the circle is
$$C_4: \left[x-\left(\frac{14}{3} + \frac 4{\sqrt 3}\right)\right]^2 + \left[y-\left(-1 - \frac{4}{\sqrt3}\right)\right]^2 = \left(\frac{14}{3} + \frac 4{\sqrt 3}\right) ^2.$$
Graphing the given circle with the $4$ possible circles, with the $2$ externally-touching circles in red, and the $2$ internally-touching circles in blue:
The equation of the circle is
$(x - p)^2 + (y - q)^2 = r^2 $
Since it touches the $y$ axis, then $ p = r $. Hence the equation becomes
$ x^2 - 2 x r + y^2 - 2 q y + q^2 = 0 $
in which there are two unknowns $q$ and $r$.
Since $M(4,3)$ is on the circle then
$ 16 - 8 r + 9 - 6 q + q^2 = 0 $
i.e.
$ 25 - 8 r - 6 q + q^2 = 0 $
And since the circle touches $(x - 2)^2 + y^2 = 1 $ , then the distance between the two centers is the sum of the radii, i.e.
$ (1 + r)^2 = (r - 2 )^2 + q^2 $
which reduces to
$ 6 r - q ^ 2 - 3 = 0 $
so that $ r = \dfrac{1}{6} ( q^2 + 3) $
Substituting this into the previous equation,
$ 25 - \dfrac{4}{3} ( q^2 + 3) - 6 q + q^2 = 0 $
i.e.
$ - \dfrac{1}{3} q^2 - 6 q + 21 = 0 $
Multiply through by $ (-3) $
$ q^2 + 18 q - 63 = 0 $
which factorizes into
$ (q + 21)(q - 3) = 0 $
The two solutions are $q = -21$ and $q = 3 $
Corresponding to these values,
$ r = \dfrac{1}{6}(q^2 + 3) = 74 $ and $ 2 $ respectively.
Therefore, the two possible circles are
$ (x - 74)^2 + (y + 21)^2 = 74^2 = 5476 $
and
$ (x - 2)^2 + (y - 3)^2 = 2^2 = 4 $
Let $O(a,b)$ the center of the wanted circle. Then its radius is equal to $a$ since it is tangent to $y$-axis. Let $O'$ be the center of $(x-2)^2+y^2=1$.Then,
The solutions are $(a,b)=(2,3)$ and $(a,b)=(74,-21).$ So we have two circles satisfying the given conditions: $(x-2)^2+(y-3)^2=4$ and $(x-74)^2+(y+21)^2=5476.$
We seek all possible circles with equations of the form $ \ (x - p)^2 + (y - q)^2 \ = \ r^2 \ $ which pass through the point $ \ (4 \ , \ 3) \ \ $ and are tangent to both the $ y-$axis and the circle $ \ (x - 2)^2 + y^2 \ = \ 1 \ \ . \ $ Since there is then a normal line at the $ \ y-$intercept of the circle(s) sought which is "horizontal", and this normal line thus contains a radius of the circle, the radius of said circle is $ \ r \ = \ p \ \ . \ $ As this tangent is "vertical" and the circle passes through a point "to the right" of the $ \ y-$axis, the circle must lie entirely in the "right half-plane". (This tangent point must therefore be $ \ (0 \ , \ q) \ \ : \ $ we can also show this by differentiating the circle equation implicitly with respect to $ \ y \ \ , \ $ producing $$ 2·(x - p) · \frac{dx}{dy} \ + \ 2·(y - q) \ \ = \ \ 0 $$ [for a vertical tangent] $$ \Rightarrow \ \ \frac{dx}{dy} \ \ = \ \ \frac{q \ - \ y}{x \ - \ p } \ \ = \ \ 0 \ \ \Rightarrow \ \ y \ = \ q \ \ , \ \ x \ \neq \ p \ \ . \ ) $$
One way in which we might ensure that we will cover all possible circles is to consider the line segment from $ \ (4 \ , \ 3) \ $ to this vertical tangent point and allow it to have any slope $ \ m \ = \ \frac{3 \ - \ q}{4 \ - \ 0 } \ \ . \ $ We then have $ \ q \ = \ 3 - 4m \ \ ; \ $ from the distance-squared between $ \ (4 \ , \ 3) \ $ and the center of the circle, we obtain $$ (4 \ - \ p)^2 \ + \ (3 \ - \ q)^2 \ \ = \ \ p^2 \ \ \Rightarrow \ \ 16 \ - \ 8p \ + \ (4m)^2 \ \ = \ \ 0 \ \ \Rightarrow \ \ 8p \ = \ 16m^2 \ + \ 16 $$ $$ \Rightarrow \ \ p \ = \ 2·(m^2 \ + \ 1) \ \ . $$ Our circle(s) may thus be described by $$ ( \ x \ - \ 2·[m^2 \ + \ 1] \ )^2 \ + \ ( \ y \ + \ 4m \ - 3 \ )^2 \ \ = \ \ 4·[m^2 \ + \ 1]^2 $$ $$ ( \ x^2 \ - \ 4·[m^2 \ + \ 1]·x \ ) \ + \ ( \ y \ + \ 4m \ - 3 \ )^2 \ \ = \ \ 0 \ \ , \quad \quad \mathbf{[ \ A \ ]} $$ for which we will be interested in the intersections of such circles with $ \ (x - 2)^2 \ + \ y^2 \ = \ 1 $ $ \rightarrow \ x^2 - 4x + y^2 + 3 \ = \ 0 \ \ \ \mathbf{[ \ B \ ]} \ . $
If we solve these as a system of equations by subtracting $ \ \mathbf{B} \ $ from $ \ \mathbf{A} \ \ , \ $ we produce a linear equation which represents the line of intersections between the circles:
$$ -4m^2·x \ + \ 2·(4m - 3)·y \ + \ (4m - 3)^2 - 3 \ \ = \ \ 0 $$ $$ \rightarrow \ \ (4m - 3)·y \ \ = \ \ 2m^2·x \ - \ (8m^2 \ - \ 12m \ + \ 3) \ \ . $$
Upon inserting this line equation into either of the circle equations, we obtain a quadratic equation for which the number of real solutions indicates the number of intersections of the circles. We shall be concerned with those values of $ \ m \ $ (if they exist) for which there is a single real solution, representing the point of tangency between the circles. It is more conventient to use $ \ (x - 2)^2 \ + \ y^2 \ = \ 1 \ \ , \ $ from which we get $$ (4m \ - \ 3)^2·( \ x^2 \ - \ 4x \ + \ 3 \ ) \ + \ [ \ 2m^2·x \ - \ (8m^2 \ - \ 12m \ + \ 3) \ ]^2 \ = \ 0 $$
$$ \rightarrow \ \ ( \ 4m^4 \ + \ 16m^2 \ - \ 24m \ + 9 \ ) · x^2 \ + \ ( \ -32m^4 \ + \ 48m^3 \ - \ 76m^2 \ + \ 96m \ - \ 36 \ )·x $$ $$ + \ ( \ 64m^4 \ - \ 192m^3 \ + \ 240m^2 \ - \ 144m \ + \ 36 \ ) \ \ = \ \ 0 \ \ . $$
The discriminant of this equation is the rather daunting $$ \Delta \ \ = \ \ -768·m^6 \ + \ 7296·m^5 \ - \ 19376·m^4 \ + \ 21120·m^3 \ - \ 10080·m^2 \ + \ 1728·m \ \ , \ $$ which (happily) has rational zeroes(!) and factors "nicely" as $$ \Delta \ \ = \ \ -16 \ · \ m \ · \ (4m \ - \ 3)^2 \ · \ (m \ - \ 6) \ · \ (3m^2 \ - \ 6m \ + \ 2) \ \ , \ $$ the quadratic factor having the real zeroes $ \ m \ = \ 1 \ \pm \ \frac{1}{\sqrt3} \ \ . $ The sign of this discriminant then tells us that the circles have
• no intersections for $ \ m \ < \ 0 \ \ \ , \ \ \ 1 - \frac{1}{\sqrt3} \ < \ m \ < \ 1 + \frac{1}{\sqrt3} \ \ \ , \ \ \ m \ > \ 6 \ \ \ $ and
• two intersections for $ \ 0 \ < \ m \ < \ 1 - \frac{1}{\sqrt3} \ \ $ and $ \ 1 + \frac{1}{\sqrt3} < \ m \ < \ 6 \ \ . $
The zeroes of the discriminant corresponds to four possible circles passing through $ \ (4 \ , \ 3) \ $ and meeting the given tangency requirements:
$ \mathbf{m \ = \ 0 \ \ : } \quad \quad ( \ x \ - \ 2 \ )^2 \ + \ ( \ y \ - \ 3 \ )^2 \ \ = \ \ 2^2 \quad $ [circles externally tangent at the uppermost point of $ \ (x - 2)^2 + y^2 \ = \ 1 \ \ $ ; the "trivial" case and probably the main one the problem-poser was after]
$ \mathbf{m \ = \ 1 - \frac{1}{\sqrt3} \ \ : } \quad \quad \left( \ x \ - \ \left[\frac{14}{3} - \frac{4}{\sqrt3} \right] \ \right)^2 \ + \ \left( \ y \ - \ \left[ \frac{4}{\sqrt3} - 1 \right] \ \right)^2 \ \ = \ \ \left[\frac{14}{3} - \frac{4}{\sqrt3} \right]^2 \ \ $ [circles are internally tangent on the lower half of the small circle]
$ \mathbf{m \ = \ 1 + \frac{1}{\sqrt3} \ \ : } \quad \quad \left( \ x \ - \ \left[\frac{14}{3} + \frac{4}{\sqrt3} \right] \ \right)^2 \ + \ \left( \ y \ + \ \left[ \frac{4}{\sqrt3} + 1 \right] \ \right)^2 \ \ = \ \ \left[\frac{14}{3} + \frac{4}{\sqrt3} \right]^2 \ \ $ [circles are internally tangent on the upper half of the small circle]
$ \mathbf{m \ = \ 6 \ \ : } \quad \quad ( \ x \ - \ 74 \ )^2 \ + \ ( \ y \ + \ 21 \ )^2 \ \ = \ \ 74^2 \quad $ [circles externally tangent on the lower half of the small circle] .
The "double zero" of the discriminant, $ \ m \ = \ \frac34 \ \ , $ corresponds to a "vertical" line of intersections "to the left" of the $ \ y-$axis and so does not give real intersections of the circles. It therefore does not affect the conclusions above.
These circles are well illustrated in the graphs in peterwhy's answer.
