I have to find the locus of centres of circles that pass through the point $(8,0)$ and tangent to circle $x^2+y^2=100$

$(x_0, y_0)$ - centres we are looking for

I decided to try something with distances, so I made a the equation $(x_0^2-8)^2+y_0^2=r^2$ It is the distance from center to point

Also I found the equation for tangent for our big circle $x_1x+y_1y-100=0$, where $x_1, y_1$ a point on the circle

Equatation for distance from center to tangent is $\frac {(x_1x_0+y_1y_0-100)^2} {x_1^2+y_1^2} =r^2$

But I dont have any results

  • $\begingroup$ The point of tangency is colinear with the centers of the two circles. $\endgroup$
    – amd
    Commented Jun 8, 2017 at 22:57
  • $\begingroup$ Interesting question (+1). Here's a desmos implementation. $\endgroup$ Commented Jun 11, 2017 at 15:08

2 Answers 2


Note that $P = (8,0)$ is located inside the circle $O: x^2 + y^2 = 10^2$, which is centered at the origin and has radius $10$. A circle tangent to $O$ and passing through $P$ consequently will be internally tangent to $O$. Suppose such a circle, which we will label $Q$, has center $(x_Q, y_Q)$; then $(x_Q, y_Q)$ is equidistant from $P$ and the point of tangency of $Q$ to $O$, and this distance is the radius of $Q$, which we will denote $r_Q$. That is to say, $$r_Q^2 = (x_Q - 8)^2 + (y_Q - 0)^2 = \left(10 - \sqrt{x_Q^2 + y_Q^2}\right)^2.$$ The rightmost expression arises from the fact that the point of tangency of two circles lies on the line joining their centers; thus the distance of $(x_Q, y_Q)$ to the point of tangency of $Q$ to $O$ is equal to the radius of $O$ minus the distance of $(x_Q, y_Q)$ to the origin.

Simplifying the RHS equality then gives $$(x_Q - 8)^2 + y_Q^2 = 10^2 - 20 \sqrt{x_Q^2 + y_Q^2} + x_Q^2 + y_Q^2,$$ or $$16x_Q + 36 = 20 \sqrt{x_Q^2 + y_Q^2},$$ or $$(4x_Q + 9)^2 = 25(x_Q^2 + y_Q^2),$$ or $$9x_Q^2 - 72x_Q + 25y_Q^2 = 81,$$ Completing the square gives $$9(x_Q - 4)^2 + 25y_Q^2 = 225,$$ or in standard form, $$\frac{(x_Q - 4)^2}{5^2} + \frac{y_Q^2}{3^2} = 1.$$ This is an ellipse with center $(4,0)$, major semiaxis $5$, and minor semiaxis $3$.

  • $\begingroup$ Worth noting the foci of this elipse are (0,8) and (0,0) $\endgroup$
    – Doug M
    Commented Jun 8, 2017 at 22:52
  • $\begingroup$ @DougM Yes indeed, yours is an important observation. In the general case, if a circle $P$ is inside another circle $O$, the locus of the centers of circles that are externally tangent to $P$ and internally tangent to $O$ is an ellipse with foci situated at the centers of $P$ and $O$. $\endgroup$
    – heropup
    Commented Jun 8, 2017 at 23:16
  • $\begingroup$ math.stackexchange.com/questions/336622/… Here $$\sqrt{(x_Q-8)^2+(y_Q-0)^2}+\sqrt{(x_Q-0)^2+(y_Q-0)^2}=10$$ $\endgroup$ Commented Jun 9, 2017 at 17:43
  • $\begingroup$ @labbhattacharjee Yes, but it is not obvious (and neither your link nor your comment explains why) that the sum of the distances of the center $(x_Q, y_Q)$ to $(0,0)$ and $(8,0)$ should be constant. The reason why is because if $r_P, r_O$ are the radii of the external and internal tangent circles, and $r_Q$ is the radius of the circle whose center draws the locus, a geometric argument shows that the sum of the distances is $$(r_P + r_Q) + (r_O - r_Q) = r_P + r_Q = \text{constant}.$$ $\endgroup$
    – heropup
    Commented Jun 9, 2017 at 18:15
  • $\begingroup$ Sorry, the previous formula should read $$(r_P + r_Q) + (r_O - r_Q) = r_P + r_O = \text{constant}.$$ $\endgroup$
    – heropup
    Commented Jun 9, 2017 at 18:21

Here we use the definition of an ellipse as a locus of points based on the director circle, which states the following:

Consider an ellipse with foci $F_1, F_2$ and major axis $2a$, and director circle $C$ with centre $F_2$ and radius $2a$. Let $P$ be a point on the ellipse. By definition, $|PF_1|+|PF_2|=2a$. Per property of the director circle, $|PF_1|=|PC|$, where |PC| is the distance from $P$ to the director circle.

Compare this with information provided in the question.

Let circle $C$ be $x^2+y^2=10^2$ with radius $2a=10$, and points $F_1=(8,0), F_2=(0,0)$.Let $P$ be the centre of the internal circle. For the internal circle to be tangential to circle $C$, and to pass through $F_1$ (as specified in the question), we require $|PC|=|PF_2|$.

Hence, by definition of the ellipse based on the director circle described above, the locus of $P$ is an ellipse with foci $F_1, F_2$ and semi-major axis $a=5$.

By symmetry, the centre of the ellipse is $(4,0)$. To find the minor axis, consider the case where $P=P'=(4,b)$. By symmetry, $F_1P=PC=PF_2=5$ and by Pythagoras' theorem, $b=3$,which is also the semi-minor axis of the ellipse.

Hence the locus of the centre of the internal circle is an ellipse with equation $$\color{red}{\frac {(x-4)^2}{5^2}+\frac {y^2}{3^2}=1}$$

See desmos implementation here.

enter image description here


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