Equation of circle touching three circles, two of which are intersecting

Question

Find the equation of the circle which is tangentially touching three given circles: $x^2+y^2=49$, $x^2+(y-3.5)^2=49/4$, and $y^2+(x-3.5)^2=49/4$.
By tangentially i mean, it touches the smaller two circle externally and the larger one internally.

The problem would have been much easier had the three given circles been tangent to each other but the smaller two of them intersect, making finding the radius of the circle in question much difficult for me.
Well, I should write the general equation of the circle and equate sum of radius with the distance between the centre with the three given equations. I am not getting the correct answer with this approach. Please help.

Answer 1

Hint: since the two small circles are symmetric to the 45° line, the midpoint of the two possible circles are on this line. one of the three possible solutions ist the large circle itself.

Answer 2

Say the center of the circle is at $(h,k)$, and it’s radius is $r$.

From the fact that it is touching the “smaller” circles, the distances between its center and their centers are equal to the sum of the radii of the two chosen circles.

This gives us the following equations:

$$\Bigl(h-\frac 72\Bigr)^2 + k^2 = \Bigl(r + \frac 72\Bigr)^2…(1)$$ $$h^2 + \Bigl(k-\frac 72\Bigr)^2 = \Bigl(r + \frac 72\Bigr)^2…(2)$$

Solving them simultaneously gives us this relation:

$$h=k…(3)$$

Now, since it touches the “large” circle internally, the distance between their centers is the difference between their radii.

So, we have

$$h^2 + k^2 = (7-r)^2…(4)$$

Solving $(2), (3) and (4)$, it is now a trivial task to find the required circles:

$$\left(x-\frac{14}{3\sqrt{2}-1}\right)^2+\left(y-\frac{14}{3\sqrt{2}-1}\right)^2=\left(\frac{7\left(\sqrt{2}-1\right)}{3\sqrt{2}-1}\right)^2$$

and

$$\left(x+\frac{14}{3\sqrt{2}+1}\right)^2+\left(y+\frac{14}{3\sqrt{2}+1}\right)^2=\left(\frac{7\left(\sqrt{2}+1\right)}{3\sqrt{2}+1}\right)^2$$

Answer 3

Let $a$ be the radius of the inscribed circle and apply the cosine rule to the triangle formed by the centers of the large circle, one of the small circles and the inscribed circle,

$$(a+3.5)^2=(7-a)^2+3.5^2-2\cdot 3.5 \cdot (7-a)\cos 135^\circ$$

which yields

$$a= \frac{7(\sqrt{2}+1)}{3\sqrt{2}+1}$$

Therefore, the equation of the inscribed circle is

$$\left( x+ \frac{7-a}{\sqrt{2}} \right)^2 +\left( y+ \frac{7-a}{\sqrt{2}} \right)^2=a^2$$

Similarly, let $b$ be the radius of the smaller inscribed circle. With the cosine rule,

$$(b+3.5)^2=(7-b)^2+3.5^2-2\cdot 3.5 \cdot (7-b)\cos 45^\circ$$

which yields

$$b= \frac{7(\sqrt{2}-1)}{3\sqrt{2}-1}$$

Therefore, the equation of the smaller inscribed circle is

$$\left( x- \frac{7-b}{\sqrt{2}} \right)^2 +\left( y- \frac{7-b}{\sqrt{2}} \right)^2=b^2$$

Answer 4

Symmetry of $3$ circles guarantee that the center of desired circle is on the bisects of first and third quadrant, that is $y=x$, so we can suppose to be $(t,t)$. Also the symmetry shows the tangent point to the circumcircle ,the big one, is on this line too. Obviously the intersections of circumcircle and line are $\pm(7/\sqrt 2,7/\sqrt 2)$. So it remains to equal the distance of center of desired circle to the perimeter of $3$ others we have already. Distance of circumcircle if clear since we have the point of tangency exactly. but for $2$ incircles, small ones, it suffice to calculate distance from center and subtract the radius. So we have: $$r_\pm=\left\|(t_\pm,t_\pm)-(\frac{7}{\sqrt 2},\frac{7}{\sqrt 2})\right\| = \sqrt 2(\frac{7}{\sqrt 2} \mp t_\pm)$$ $$(t_\pm-\frac{7}{2})^2+(t_\pm-0)^2 = ((7 \mp \sqrt 2t_\pm) + \frac{7}{2})^2 \Longrightarrow t_\pm=\frac{\pm14}{3\sqrt 2 \mp 1}$$ $$r_\pm = 7 - \frac{14}{3\sqrt 2 \mp 1}$$ $$(x-t_\pm)^2+(y-t_\pm)^2=r_\pm^2$$ That up indices used for smaller circle that placed in positive area,

And down indices for bigger circle that placed in negative area.