Equation of a circle that passes through $(4,3)$ and which, touches the $y$ axis and another given circle

1. Find the equation of a circle which passes through $$M(4,3)$$, touches the $$y$$ axis and touches the circle $$(x-2)^2 + y^2 =1$$

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?

• As the required circle touches the $y$-axis, the $x$-coordinate of its centre is related to $r$. Commented Mar 18, 2023 at 23:01
• @peterwhy Thanks for the comment! So is $p=r$? What about the point $M(4,3)$? Commented Mar 18, 2023 at 23:04
• If you know how to find the new circle which touches a given circle, you can treat the point $(4,3)$ as a circle with zero radius. Commented Mar 18, 2023 at 23:07
• @peterwhy Can you tell me if this is correct: $(x-2)^2 +(y-3)^2=4$? Commented Mar 18, 2023 at 23:21
• Your circle seems to satisfy the $3$ conditions if my calculation is right. I am also checking if there's another possible circle which touches the given circle internally. (Even though the question only asks for one circle) Commented Mar 18, 2023 at 23:28

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,

1. $$OO'=a+1\implies (a-2)^2+b^2=(a+1)^2$$
2. $$OM=a\implies (a-4)^2+(b-3)^2=a^2$$

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.